Following the comments, I post here some progress on the 2nd identity. The main results are (1) and (2), and it remains to prove (3). Probably I will finish the calculation later or set a bounty if nobody comes up with a simpler idea.
Second identity: Using the same notation as above, we can rewrite the integral as
$$ A=\int_0^{\pi}\left(\int_0^{\pi}\ln I(x,y)\;dy\right)dx=\left(\int_0^{\frac{2\pi}{3}}+\int_{\frac{2\pi}{3}}^{\pi}\right)\left(\int_0^{\pi}\ln I(x,y)\;dy\right)dx.$$
-
For $x\in(0,2\pi/3)$ we have $|2\cos\frac{x}{2}|>1$ and can write (using (2) from the above)
\begin{align}
F_+(x)=\int_0^{\pi}\ln I(x,y)\;dy=\\=\int_0^{\pi}\left[\ln\left(1+\frac{1}{2\cos\frac{x}{2}}e^{-i(y-x/2)}\right)+\ln\left(1+\frac{1}{2\cos\frac{x}{2}}e^{i(y-x/2)}\right)+\ln 4\cos^2\frac{x}{2}\right]dy=\\
=i\left[\mathrm{Li}_2\left(\frac{e^{-ix/2}}{2\cos\frac{x}{2}}\right)-\mathrm{Li}_2\left(-\frac{e^{-ix/2}}{2\cos\frac{x}{2}}\right)+\mathrm{Li}_2\left(-\frac{e^{ix/2}}{2\cos\frac{x}{2}}\right)-\mathrm{Li}_2\left(\frac{e^{ix/2}}{2\cos\frac{x}{2}}\right)\right]+\\+\pi \ln 4\cos^2\frac{x}{2}.
\end{align}
At the last step we have used the standard evaluations of $\mathrm{Li}_2(z)$:
\begin{align}
\int_0^{\pi}\ln\left(1+re^{\pm iy}\right)dy=\pm i\Bigl[\mathrm{Li}_2(r)-\mathrm{Li}_2(-r)\Bigr],\qquad |r|<1.
\end{align}
- Similarly, for $x\in(2\pi/3,\pi)$ we have $|2\cos\frac{x}{2}|<1$, so that
\begin{align}
F_-(x)=\int_0^{\pi}\ln I(x,y)\;dy=\qquad\\=\int_0^{\pi}\left[\ln\left(1+2\cos\frac{x}{2}e^{-i(y-x/2)}\right)+\ln\left(1+2\cos\frac{x}{2}e^{i(y-x/2)}\right)\right]dy=\qquad\\
=i\left[\mathrm{Li}_2\left(2\cos\frac{x}{2}e^{-ix/2}\right)-\mathrm{Li}_2\left(-2\cos\frac{x}{2}e^{-ix/2}\right)+\mathrm{Li}_2\left(-2\cos\frac{x}{2}e^{ix/2}\right)-\mathrm{Li}_2\left(2\cos\frac{x}{2}e^{ix/2}\right)\right].
\end{align}
$$=i\left[\mathrm{Li}_2\left(1+e^{-ix}\right)-\mathrm{Li}_2\left(1+e^{ix}\right)-\mathrm{Li}_2\left(-1-e^{-ix}\right)+\mathrm{Li}_2\left(-1-e^{ix}\right)\right].$$
The functions $F_+(x)$ and $F_-(x)$ can be analytically continued outside their respective domains of definition. It turns out that these continuations coincide in an open neibourhood of the real interval $x\in(0,\pi)$. This is after all not very surprising since both $F_{\pm}(x)$ represent the same integral $\int_0^{\pi}\ln I(x,y)\;dy$. In order to show that, indeed, $F_+(x)=F_-(x)$ for $x\in(0,\pi)$, one should use the identities of type $\mathrm{Li}_2(z)+\mathrm{Li}_2(z^{-1})=-\frac{\pi^2}{6}-\frac12 \ln^2(-z)$. [Or simply verify that the derivatives (given by elementary functions) of $F_{\pm}(x)$ coincide and compute both functions at some particular point].
Now, since $F_+(x)=F_-(x)$ on $x\in(0,\pi)$, we can write
\begin{align}
A=\int_0^{\frac{2\pi}{3}}F_+(x)dx+\int_{\frac{2\pi}{3}}^{\pi}F_-(x)dx=\int_0^{\pi}F_-(x)dx=\\=
i\int_0^{\pi}\left[\mathrm{Li}_2\left(1+e^{-ix}\right)-\mathrm{Li}_2\left(1+e^{ix}\right)-\mathrm{Li}_2\left(-1-e^{-ix}\right)+\mathrm{Li}_2\left(-1-e^{ix}\right)\right]\tag{1}dx.
\end{align}
N.B.: Dilogarithms $\mathrm{Li}_2(z)$ are defined on their main sheet: the cut plane $z\in\mathbb{C}\backslash[1,\infty)$.
Let us show how to compute the first two integrals in (1). This can be done as follows. First, use the
identity $\mathrm{Li}_2(z)+\mathrm{Li}_2(1-z)=-\ln z\ln(1-z)+\frac{\pi^2}{6}$ to rewrite these integrals as
\begin{align}
A_1=i\int_0^{\pi}\left[\mathrm{Li}_2\left(1+e^{-ix}\right)-\mathrm{Li}_2\left(1+e^{ix}\right)\right]dx=\\=
i\int_0^{\pi}\left[\mathrm{Li}_2\left(-e^{ix}\right)-\mathrm{Li}_2\left(-e^{-ix}\right)-i(\pi-x)\ln4\cos^2\frac{x}{2}\right]dx.
\end{align}
For the first two terms, introduce $z=e^{ix}$ and deform the demi-circles of integration to diameters. This yields
$$i\int_0^{\pi}\left[\mathrm{Li}_2\left(-e^{ix}\right)-\mathrm{Li}_2\left(-e^{-ix}\right)\right]dx=2\int_{-1}^{1}\frac{\mathrm{Li}_2(z)}{z}dz=
\frac{7}{2}\zeta(3),$$
where the last equality is obtained by expanding $\mathrm{Li}_2(z)$ into series, computing the integrals and using that
$\zeta(3)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^3}+\frac{1}{8}\zeta(3)$. Next, in $\int_0^{\pi}(\pi-x)\ln4\cos^2\frac{x}{2}\,dx$,
integrating once by parts to kill the linear term, we obtain one half of the initial integral $A_1$. Therefore,
$$A_1=i\int_0^{\pi}\left[\mathrm{Li}_2\left(1+e^{-ix}\right)-\mathrm{Li}_2\left(1+e^{ix}\right)\right]dx=7\zeta(3)\tag{2}$$
and now it remains to show that
$$A_2=i\int_0^{\pi}\left[\mathrm{Li}_2\left(-1-e^{ix}\right)-\mathrm{Li}_2\left(-1-e^{-ix}\right)\right]dx=\frac{7}{3}\zeta(3).\tag{3}
$$
