I present a different solution utilising elementary real analytic techniques, namely Mellin transforms.
Consider the following classical Mellin transform $$\int_{0}^{\infty} \frac{x^{s-1}}{e^{n x}+1} \, dx \stackrel{x \mapsto u/n}{=} n^{-s} \int_{0}^{\infty} \frac{u^{s-1}}{e^u+1}\, du$$
By writing $$\frac{1}{e^u+1} = \frac{e^{-u}}{1+e^{-u}} = -\sum_{k=1}^{\infty} (-e^u)^{-k}=\sum_{k=1}^{\infty} (-1)^{k-1} e^{-k u}$$
we have
\begin{align*}\int_{0}^{\infty} \frac{u^{s-1}}{e^u+1}\, du &= \sum_{k=1}^{\infty} (-1)^{k-1} \int_{0}^{\infty} u^{s-1}e^{-k u}\, du\\&\stackrel{u \mapsto t/k}=\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^s} \int_{0}^{\infty} t^{s-1} e^{-t} \, dt \\&= \eta (s)\Gamma (s)=(1-2^{1-s})\zeta(s) \Gamma(s)\end{align*}
Thus $$\int_{0}^{\infty} \frac{x^{s-1}}{e^{n x}+1} \, dx = n^{-s} (1-2^{1-s})\zeta(s) \Gamma(s)$$ for $\Re(n)>0 \land \Re (s)>0$.
Now differentiate this Mellin transform with respect to $n$ to obtain $$\int_{0}^{\infty} \frac{x^{s}e^{n x}}{(1+e^{n x})^2}\,dx = s (1-2^{1-s})n^{-1-s} \zeta(s) \Gamma (s)$$
Take $n \mapsto 2$ to obtain $$\int_{0}^{\infty} \frac{x^s e^{2 x}}{(1+e^{2x})^2} \,dx=s\, 2^{-1-s} (1-2^{1-s}) \zeta (s) \Gamma (s)\tag{1}$$
Now similarly consider the following Mellin transform $$\int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{e^{n x}+1} \, dx \stackrel{x \mapsto u/n}{=} n^{-s} \int_{0}^{\infty} \frac{u^{s-1}e^{-u/n}}{e^u+1}\, du$$
By once again writing $$\frac{1}{e^u+1} = \frac{e^{-u}}{1+e^{-u}} = -\sum_{k=1}^{\infty} (-e^u)^{-k}=\sum_{k=1}^{\infty} (-1)^{k-1} e^{-k u}$$
we have \begin{align*}\int_{0}^{\infty} \frac{u^{s-1}e^{-u/n}}{e^u+1}\, du &= \sum_{k=1}^{\infty} (-1)^{k-1} \int_{0}^{\infty} u^{s-1} e^{-u(k+1/n)}\, du\\&=\Gamma (s) \sum_{k=1}^{\infty} (-1)^{k-1} \left(k+\frac{1}{n}\right)^{-s}\end{align*} Separating this alternating sum into odd and even terms, we can conclude $$\int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{e^{n x}+1} \, dx = 2^{-s} n^{-s} \left(\zeta\left(s, \frac{1+n}{2n}\right)-\zeta \left(s, \frac{1+2n}{2n} \right)\right)\Gamma (s)$$
for $\Re (n)>0 \land \Re(s)>0$ where $\zeta(s,a)$ is the Hurwitz zeta function.
Once again differentiate with respect to $n$ and take $n \mapsto 2$ to obtain
$$-\int_{0}^{\infty} \frac{x^s e^x}{(1+e^{2 x})^2}\,dx=s\, 2^{-3-2s}\left(4\zeta \left(s,\frac{5}{4}\right)-4\zeta\left(s,\frac{3}{4}\right)+\zeta \left(1+s,\frac{3}{4}\right)-\zeta \left(1+s,\frac{5}{4}\right)\right)\Gamma(s)\tag{2}$$
Proceed to add equations $(1)$ and $(2)$ to obtain $$\int_{0}^{\infty} \frac{x^s (e^{2x}-e^x)}{(1+e^{2x})^2}\,dx=2^{-2 s-3} s \Gamma (s) \left(4 \left(2^s-2\right) \zeta (s)+4 \zeta \left(s,\frac{1}{4}\right)-4 \zeta \left(s,\frac{3}{4}\right)-\zeta \left(s+1,\frac{1}{4}\right)+\zeta\left(s+1,\frac{3}{4}\right)\right)$$
or upon shifting $s \mapsto s-1$
$$\int_{0}^{\infty} \frac{x^{s-1} (e^{2x}-e^x)}{(1+e^{2x})^2}\,dx=\frac{1}{2}\Gamma (s) \left(2^{1-s}\eta(s-1)+\beta(s-1)-\beta(s)\right)\tag{3}$$
where $\eta$ and $\beta$ are the Dirichlet eta and beta functions respectively.
(This next limit swapping bit needs to be rigorously shown to be allowed)
Now taking the limit of $(3)$ as $s \to 0$, we obtain $$\int_{0}^{\infty} \frac{e^{2x}-e^x}{x(1+e^{2x})^2} = -\frac{1}{4}+\frac{G}{\pi}-\frac{1}{3} \ln(2)-\frac{1}{2}\ln (\pi)+3\ln (A)+\ln \Gamma \left(\frac{3}{4}\right)$$ as desired based on known values of the derivative of the Dirichlet eta and beta functions. $\square$
