Evaluation of $\int_{0}^{1}{\int_{0}^{1}{\frac{\ln \left( 1-x \right)-\ln \left( 1-y \right)}{x-y-1}dx}dy}$

This is not a complete answer, but too long for a comment. I managed to reduce the two dimensional integral into a single definite integral

$$\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)-\ln\left(1-y\right)}{x-y-1}{\rm d}x{\rm d}y=\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)}{x-y-1}{\rm d}x{\rm d}y-\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-y\right)}{x-y-1}{\rm d}x{\rm d}y=$$

$$=\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)}{x-y-1}{\rm d}x{\rm d}y-\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)}{y-x-1}{\rm d}x{\rm d}y=$$

$$=\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)}{x-y-1}{\rm d}x{\rm d}y+\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)}{x-y+1}{\rm d}x{\rm d}y=$$

$$=\int_{0}^{1}\ln\left(1-x\right)\left(\ln\left(1-x\right)-\ln\left(2-x\right)\right){\rm d}x+\int_{0}^{1}\ln\left(1-x\right)\left(\ln\left(1+x\right)-\ln x\right){\rm d}x=$$

$$=\int_{0}^{1}\ln\left(1-x\right)\left(\ln\left(1-x\right)-\ln\left(2-x\right)+\ln\left(1+x\right)-\ln x\right){\rm d}x=$$

$$=\int_{0}^{1}\ln\left(1-x\right)\ln\left(\dfrac{1-x^{2}}{2x-x^{2}}\right){\rm d}x$$

According to Wolfram this integral involves the polylogarithm ${\rm Li}_{2}\left(x\right)$ and can be evaluated. Note that

$${\rm Li}_{2}\left(x\right)=\sum_{n=1}^{\infty}\dfrac{x^{n}}{n^2}$$

so you can immediately see the connection to ${\rm Li}_{2}\left(1\right)=\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$.

Performing the change of variable $$ u=\frac{1-x}{1-y}\,\,,\,\,v=x-y$$ the domain $D=\{(x,y)\,|\,0<x<1\,,\,0<y<1\}$ change into $$D_1=\{(u,v)\,|\,0<u<1\,,\,0<v<1-u\}\cup D_2=\{(u,v)\,|\,1<u<\infty\,,\,0<v<\frac{1-u}{u}\}$$

and your integral reads

$$\int_{0}^{1}\int_{0}^{1}\dfrac{\ln\left(1-x\right)-\ln\left(1-y\right)}{x-y-1}{\rm d}x{\rm d}y=$$ $$\int_{0}^{1}\int_{0}^{1-u}\dfrac{\ln\left(u\right)}{v-1}\frac{v}{(u-1)^2}{\rm d}v{\rm d}u-\int_{1}^{\infty}\int_{\frac{1-u}{u}}^0\dfrac{\ln\left(u\right)}{v-1}\frac{v}{(u-1)^2}{\rm d}v{\rm d}u=$$

Here, the negative sign becomes from de Jocobian. The first integral in $v$ (in both cases) is inmediatly and get

$$\int_0^1 \frac{(1-u+\log (u)) \log (u)}{(u-1)^2} \, du -\int_1^{\infty } \frac{\left(1-\frac{1}{u}-\log \left(2-\frac{1}{u}\right)\right) \log (u)}{(u-1)^2} \, du$$

and with the change $1/u\rightarrow u$ in the second integral we arrives to

$$\int_0^1 \frac{(2(1-u)+\log (u)-\log(2-u)) \log (u)}{(u-1)^2} \, du$$

Now:

• $\displaystyle \int_0^1 \frac{2(1-u)\log (u)}{(u-1)^2} \, du=-2\int_0^1 \frac{ \log (u)}{u-1} \, du=2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\int_0^1(u-1)^{n-1}du=2\sum_{n=1}^{\infty}\frac{(-1)}{n^2}=\boxed{-\frac{\pi^2}{3}}$

• Integrating by parts we obtain $\displaystyle \int_0^1 \frac{2\log (u)}{u-1} \, du=\int_0^1 \frac{\log^2 (u)}{(u-1)^2} \, du=\boxed{\frac{\pi ^2}{3}}$

Using similar techniques (Taylor series of $\log(u)$ or $\log(2-u)$) you can arrive - $\displaystyle \int_0^1 -\frac{\log (2-u)\log (u)}{(u-1)^2} \, du=\boxed{\frac{\pi ^2}{12}+\log^2(2)}$

and the desired result follows.