Note that the integrand is absolutely integrable on $(0,\infty)$; this claim will let us do things like swap sums and integrals, and break up the region of integration.
Start by writing part of the integrand using a geometric series:
\begin{align}
I=\int_{0}^{\infty} \frac{\log(1+\cos(x))}{1+e^x}\,dx &= \int_{0}^{\infty}\log(1+\cos(x))\cdot \frac{1}{1-(-e^{-x})}\cdot e^{-x}\,dx\\&=\sum_{n=1}^{\infty}(-1)^{n+1}\int _0^{\infty} \log(1+\cos(x)) e^{-nx}\,dx\end{align}
Split up $(0,\infty)$ into blocks of length $2\pi$:
$$
I=\sum_{n=1}^{\infty}(-1)^{n+1}\sum_{m=0}^{\infty}\int _{2\pi m}^{2\pi(m+1)} \log(1+\cos(x)) e^{-nx}\,dx
$$
Change of variable to standardize the new region of integration to $(0,2\pi)$:
$$
I=\sum_{n=1}^{\infty}(-1)^{n+1}\sum_{m=0}^{\infty}e^{-2mn\pi}\int _{0}^{2\pi} \log(1+\cos(x)) e^{-nx}\,dx
$$This middle sum is a geometric series:
$$
I=\sum_{n=1}^{\infty}(-1)^{n+1}\cdot\frac{1}{2}\left(1+\coth
(\pi n)\right) \cdot \int _{0}^{2\pi} \log(1+\cos(x)) e^{-nx}\,dx
$$Invoke the Fourier series for the log term on $(0,2\pi)$; this is the same as the Fourier series for $\log(1-\cos(x))$ on $(-\pi,\pi)$, if you'd prefer:
\begin{align}I&=\sum_{n=1}^{\infty}(-1)^{n+1}\cdot\frac{1}{2}\left(1+\coth
(\pi n)\right) \cdot \int _{0}^{2\pi}\left(-\log(2)+ \sum_{k=1}^{\infty}\frac{2\cos(k x)(-1)^{k+1}}{k}\right) e^{-nx}\,dx
\\&=\small\sum_{n=1}^{\infty}(-1)^{n+1}\cdot\frac{1}{2}\left(1+\coth
(\pi n)\right) \cdot\left( \int _{0}^{2\pi}-\log(2)e^{-nx}\,dx+\sum_{k=1}^{\infty} \frac{2(-1)^{k+1}}{k} \int _{0}^{2\pi}\cos(k x)e^{-nx}\,dx\right)\end{align} These new integrals are easily evaluated either directly or by integration by parts twice, and in fact cancel with the term from the middle sum:
\begin{align}
I&=\sum_{n=1}^{\infty}(-1)^{n+1}\left( \frac{-\log(2)}{n} + \sum_{k=1}^{\infty}\frac{2(-1)^{k+1}}{k}\cdot\frac{n}{n^2+k^2}\right)
\\&=\log(2)\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} + \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{2(-1)^{k+n}}{k}\cdot\frac{n}{n^2+k^2}
\end{align} But the second sum can now be evaluated using the symmetry trick from the linked question and the value of the alternating harmonic series:
\begin{align}
I&=\log(2)\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} + \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k+n}}{nk}
\\&=\log(2)\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} - \log(2)\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\\&=0
\end{align}
Some remarks. My first thought was to use $\int_{-a}^a f(x)/(1+e^x)\,dx = \int_0^a f(x)\,dx$ if $f$ is even and integrable, but the shoe's on the wrong foot, so to speak. I also tried the residue theorem and some other substitutions but they didn't pan out. It was interesting to note the prevalence of $\log(2)^2$ throughout this problem; I tried (in vain!) to connect the integrals in this problem with Putnam 2017 B4, though perhaps there is a link. As mentioned in my comment, much of this follows from your linked question. Hope this is clear and helpful.
