By the help of Mathematica numeral calculations, I find the following formula holds
$$\sum\limits_{n=1}^\infty \frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n=m\log\left(\frac{m}{m-1}\right)\quad ?$$
$m>1$ is a positive integer. But I can't prove it.
Too long for comments.
Using another CAS, I have not been able to obtain the rhs (except for $m=2$) but numerically the results do agree with your conjecture (checked up to $m=20$).
Considering $$f_m=\sum\limits_{n=1}^\infty \frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n$$ running cases, what I obtained is $$f_3=\frac{2 ^2}{3^2} \, _4F_3\left(1,1,\frac{4}{3},\frac{5}{3};\frac{3}{2},2,2;1\right)$$ $$f_4=\frac{3^3}{4^3} \, _5F_4\left(1,1,\frac{5}{4},\frac{6}{4},\frac{7}{4};\frac{4}{3},\frac{5}{3},2,2;1 \right)$$ $$f_5=\frac{4^4}{5^4} \, _6F_5\left(1,1,\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5};\frac{5}{4},\frac {6}{4},\frac{7}{4},2,2;1\right)$$ $$f_6=\frac{5^5}{6^5}\, _7F_6\left(1,1,\frac{7}{6},\frac{8}{6},\frac{9}{6},\frac{10}{6},\frac{11}{6};\frac {6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5},2,2;1\right)$$ $$f_7=\frac{6^6}{7^6}\, _8F_7\left(1,1,\frac{8}{7},\frac{9}{7},\frac{10}{7},\frac{11}{7},\frac{12}{7}, \frac{13}{7};\frac{7}{6},\frac{8}{6},\frac{9}{6},\frac{10}{6},\frac{11}{6},2,2;1 \right)$$ which, as written, reveal very clear patterns. $$\color{blue}{f_m=\frac{(m-1)^{m-1}}{m^{m-1}}\, _{m+1}F_m\left(1,1,\frac{m+1}m,\cdots,\frac{2m-1}m;\frac m{m-1},\cdots,\frac {2m-3}{m-1},2,2;1\right)}$$
Trying on Wolfram Cloud, I obtained the same results but no simplification at all. Surprising, isn't it ?
Let $z_m=(m-1)^{m-1}/m^m$. From this answer, we have \begin{align} F_m(z)&:=\sum_{n=0}^{\infty}\binom{mn}{n}\frac{z^n}{(m-1)n+1}=1+z\big(F_m(z)\big)^m, \\ G_m(z)&:=\sum_{n=0}^{\infty}\binom{mn}{n}z^n=\frac{F_m(z)}{m-(m-1)F_m(z)}. \end{align} Now $F_m(0)=1$ and $\color{blue}{F_m(z_m)=m/(m-1)}$ (yeah!), thus $$\sum_{n=1}^{\infty}\binom{mn}{n}\frac{(z_m)^n}{n}=\int_{0}^{z_m}\frac{G_m(z)-1}{z}\,dz,$$ and the substitution $w=F_m(z)$ (i.e. $z=(w-1)/w^m$) collapses it to $$\sum_{n=1}^{\infty}\binom{mn}{n}\frac{(z_m)^n}{n}=\color{blue}{m\int_1^{m/(m-1)}\frac{dw}{w}}=m\ln\frac{m}{m-1}.$$ [As a by-product, we get $\displaystyle\sum_{n=1}^{\infty}\binom{mn}{n}\frac{z^n}{n}=m\ln F_m(z)$.]
I'm posting this as another "answer", because it might be relevant, but it's not related to my original attempt. This doesn't tell how to prove the closed form, it's just an illustration of some interesting consequences of the conjecture.
In the following paper the author derives a general asymptotic series for binomial coefficients. For the case we are interested in it looks like this:
$$\binom{mn}{n} \asymp \sqrt{\frac{m}{2 \pi (m-1) n}} \left( \frac{m^m}{(m-1)^{m-1}} \right)^n \sum_{k=0}^\infty \frac{P_k(m)}{n^k} \\ n \to \infty$$
Where $$P_0(m)=1 \\ P_k(m) = \frac{1}{k} \sum_{j=1}^k \frac{(-1)^j}{j+1} \left(1+\frac{1}{(m-1)^j}-\frac{1}{m^j} \right) B_{j+1}(1) P_{k-j}(m)$$
Where $B_{j+1}(x)$ are Bernoulli polynomials.
Obviously, we can see that the first term of this asymptotic expansion exactly matches the "weird" part of the original series, so we can make another conjecture:
$$m\log\left(\frac{m}{m-1}\right) \approx \sqrt{\frac{m}{2 \pi (m-1)}} \sum_{n=1}^\infty \frac{1}{n^{3/2}} \sum_{k=0}^K \frac{P_k(m)}{n^k}$$
Where $K$ is some large but finite number. (Remember, the $k$ series is asymptotic series, it doesn't converge). So we can exchange the order of summation:
$$m\log\left(\frac{m}{m-1}\right) \approx \sqrt{\frac{m}{2 \pi (m-1)}} \sum_{k=0}^K \zeta \left(k+\frac{3}{2} \right) P_k(m)$$
Checking numerically, I found that $K=6$ or $K=8$ gives the best result for all $m \geq 2$:
Obviously, to achieve better agreement, we need to pick larger $n$, so it makes sense to write:
$$m\log\left(\frac{m}{m-1}\right) \approx \sum_{n=1}^N \binom{n m}{n} \frac{1}{n} \left( \frac{(m-1)^{m-1}}{m^m} \right)^n+ \sqrt{\frac{m}{2 \pi (m-1)}} \sum_{n=N+1}^\infty \frac{1}{n^{3/2}} \sum_{k=0}^K \frac{P_k(m)}{n^k}$$
Or:
$$m\log\left(\frac{m}{m-1}\right) \approx \sum_{n=1}^N \binom{n m}{n} \frac{1}{n} \left( \frac{(m-1)^{m-1}}{m^m} \right)^n+ \\ + \sqrt{\frac{m}{2 \pi (m-1)}} \sum_{k=0}^K \left(\zeta \left(k+\frac{3}{2} \right)-\sum_{n=1}^N \frac{1}{n^{k+3/2}} \right) P_k(m)$$
This drastically improves the accuracy, see for example $N=5$:
For $N=5$ and $K=25$, and calling the approximation $S(m)$, we have:
$$\begin{array}(m & m\log\left(\frac{m}{m-1}\right) & S(m) \\ 2 & 1.3862943611198906 & 1.3862943611198906 \\ 3 & 1.216395324324493145 & 1.216395324324493145 \\ 4 & 1.150728289807123709 & 1.150728289807123709 \\ 5 & 1.115717756571048778 & 1.115717756571048778 \\ \pi & 1.20379579648763820 & 1.20379579648763820 \end{array}$$
Where only the correct digits are shown. As you can see by the last example, irrational $m$ work just as fine.
This is hardly a starter, but the following representation showing some commonalities of LHS and RHS might be useful.
The RHS can be written as \begin{align*} m\log\left(\frac{m}{m-1}\right)&=m\log\left(\frac{1}{1-\frac{1}{m}}\right)\\ &=-m\log\left(1-\frac{1}{m}\right)\\ &\,\,\color{blue}{=m\sum_{n=1}^\infty \frac{1}{nm^n}} \end{align*}
The LHS can be written as \begin{align*} \sum_{n=1}^\infty&\frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n\\ &=\sum_{n=1}^\infty\frac{1}{nm^n}\binom{mn}{n}\left(\frac{(m-1)^{m-1}}{m^{m-1}}\right)^n\\ &=\sum_{n=1}^\infty\frac{1}{nm^n}\binom{mn}{n}\left(1-\frac{1}{m}\right)^{n(m-1)}\\ &\,\,\color{blue}{=m\sum_{n=1}^\infty\frac{1}{nm^n}\binom{mn-1}{n-1}\left(1-\frac{1}{m}\right)^{n(m-1)}} \end{align*}
Here's my attempt at closed form. Not an answer definitely, but might be of use.
First, we simplify (and generalize) the problem by defining a two variable series:
$$S(x,y)=\sum_{n=1}^\infty \binom{nx}{n} \frac{y^n}{n}$$
In the OP we have:
$$y=\frac{1}{x} \left(1-\frac{1}{x} \right)^{x-1}$$
Now we assume $x \in \mathbb{R}$, but $x \notin \mathbb{Z}$ (we can get back to the whole numbers by continuity arguments), then we can represent the binomial coefficient in the following way:
$$\binom{nx}{n}= \frac{x}{\pi}\sin[\pi n (1-x)] B[nx,n(1-x)] $$
Here's the main problem: all the (real) integral representations of the Beta function rely on both the arguments being positive. But it would only be the case if $0<x<1$, which doesn't fit the OP. However, I have checked the original series, and despite giving complex values, the closed form still seems to work for $|x|<1$, so I will consider this case first.
Attempt 1
We have:
$$B[nx,n(1-x)]=\int_0^1 t^{n x-1} (1-t)^{n(1-x)-1}dt=\int_0^1 \left[t^x (1-t)^{1-x} \right]^n \frac{dt}{t(1-t)}$$
$$\sin[\pi n (1-x)]=\frac{1}{2i} \left(e^{\pi i (1-x) n}-e^{-\pi i (1-x) n} \right)$$
Then we can write:
$$S(x,y)=\frac{x}{2 i \pi} \int_0^1 \frac{dt}{t(1-t)} \sum_{n=1}^\infty \left(e^{\pi i (1-x) n}-e^{-\pi i (1-x) n} \right) \left[t^x (1-t)^{1-x} \right]^n \frac{y^n}{n} $$
We also need $|y|<1$, which doesn't seem to work for $|x|<1$ if we define $y$ as in the original series, however let's forget about that for now and sum the series formally:
$$S(x,y)=-\frac{x}{2 \pi i} \int_0^1 \frac{dt}{t(1-t)} \log \frac{1-e^{\pi i (1-x)} t^x (1-t)^{1-x} y }{1-e^{-\pi i (1-x)} t^x (1-t)^{1-x} y}, \qquad 0<x<1$$
If we set $y=\frac{1}{x} \left(1-\frac{1}{x} \right)^{x-1}$, then the closed form $-x \log \left( 1-\frac{1}{x} \right)$ works numerically, as in, the real and imaginary parts are the same. Though I don't know how to prove it for the integral either.
Attempt 2
For another try we could turn to Gamma functions, which are better defined:
$$\binom{nx}{n}= \frac{x}{\pi}\sin[\pi n (1-x)] \frac{\Gamma(nx) \Gamma(n(1-x))}{(n-1)!} $$
To work with the usual integral representation of the Gamma function, we again have to restrict ourselves to $0 <x <1$, however, as we will see, it will allow us to consider $|y|>1 $ as well.
$$\Gamma(nx) \Gamma(n(1-x))=\int_0^\infty \int_0^\infty u^{nx} v^{n(1-x)} e^{-u-v} \frac{du dv}{u v}$$
So we have:
$$S(x,y)=\frac{x}{2\pi i} \int_0^\infty \int_0^\infty e^{-u-v} \frac{du dv}{u v} \sum_{n=1}^\infty \left(e^{\pi i (1-x) n}-e^{-\pi i (1-x) n} \right) [u^x v^{1-x}]^n \frac{y^n}{n!}$$
Summation gives us:
$$S(x,y)=\frac{x}{2\pi i} \int_0^\infty \int_0^\infty e^{-u-v} \frac{du dv}{u v} \left(\exp \left[y e^{\pi i (1-x)} u^x v^{1-x} \right]-\exp \left[y e^{-\pi i (1-x)} u^x v^{1-x} \right] \right) $$
Getting rid of complex numbers:
$$S(x,y)=\frac{x}{\pi} \int_0^\infty \int_0^\infty e^{-u-v} \exp \left[y \cos (\pi (1-x)) u^x v^{1-x} \right] \sin \left[y \sin (\pi (1-x)) u^x v^{1-x} \right] \frac{du dv}{u v}$$
This integral seems to work as well, though numerical evaluation is very difficult.
EDIT: This answer is incorrect, due to a mistaken bound on the binomial coefficient. In fact, $\binom{mn}{n}\leq 2^{mn}$ so of course it has exponential growth...
The formula cannot hold, since the left side is a divergent series! Indeed, for every fixed $m$ the $\binom{mn}{n}$ grows faster than exponentially in $n$, since it is greater than $((m-1)n)^n$. But this means it outgrows the reciprocal of the rest of the summand...
