Proving Abel's identity for the Dilogarithm.

This is a direct translation of Abel's nice paper 'Note sur la fonction $\psi x=x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+\cdots$'.

Start with $\;\displaystyle\operatorname{Li}_2(x)=-\int\frac{\log(1-x)}x\,dx\;$ and set $\;\displaystyle x:=\frac a{1-a}\frac y{1-y}\;$ with '$a$' constant.

Then $\;\log(x)=\log(a)-\log(1-a)+\log(y)-\log(1-y)\,$ and the differentials will verify : $$\frac {dx}x=\frac {dy}y+\frac {dy}{1-y}$$ Since $\,(1-a)(1-y)-ay=1-a-y\,$ we get :
(factorizing further $(1-y)$ in the logarithm for $\frac{dy}y\;$ and $(1-a)$ for $\frac{dy}{1-y}$)

\begin{align} \operatorname{Li}_2\left(\frac a{1-a}\frac y{1-y}\right)&=-\int\left(\frac{dy}y+\frac{dy}{1-y}\right)\log\frac{1-a-y}{(1-a)(1-y)}\\ &=-\int\frac{dy}y\log\left(1-\frac y{1-a}\right)+\int \frac{dy}y\log(1-y)\\ &\quad-\int\frac{dy}{1-y}\log\left(1-\frac a{1-y}\right)+\int \frac{dy}{1-y}\log(1-a)\\ \end{align}

But the integrals at the right may be written as $\operatorname{Li}_2$ functions since : \begin{align} &\int\frac{dy}y\log\left(1-\frac y{1-a}\right)=-\operatorname{Li}_2\left(\frac y{1-a}\right),\\ &\int\frac{dy}y\log\left(1-y\right)=-\operatorname{Li}_2\left(y\right);\\ \end{align} so that $$ \operatorname{Li}_2\left(\frac a{1-a}\frac y{1-y}\right)=\operatorname{Li}_2\left(\frac y{1-a}\right)-\operatorname{Li}_2\left(y\right)-\log(1-a)\log(1-y)-\int\frac{dy}{1-y}\log\left(1-\frac a{1-y}\right)$$

Set $z:=\dfrac a{1-y}$ i.e. $1-y=\dfrac az,\;dy=\dfrac {a\,dz}{z^2}$ to get : $$\int\frac{dy}{1-y}\log\left(1-\frac a{1-y}\right)=\int\frac{dz}z\log(1-z)=-\operatorname{Li}_2\left(z\right)=-\operatorname{Li}_2\left(\frac a{1-y}\right)$$

$$\operatorname{Li}_2\left(\frac a{1-a}\frac y{1-y}\right)=\operatorname{Li}_2\left(\frac y{1-a}\right)+\operatorname{Li}_2\left(\frac a{1-y}\right)-\operatorname{Li}_2\left(y\right)-\log(1-a)\log(1-y)+C$$

The arbitrary constant $C$ will be determined by $\,y=0$ to get $\,C=-\operatorname{Li}_2(a)$.
Replacing $a$ by $x$ we get :

$$\operatorname{Li}_2\left(\frac x{1-x}\frac y{1-y}\right)=\operatorname{Li}_2\left(\frac y{1-x}\right)+\operatorname{Li}_2\left(\frac x{1-y}\right)-\operatorname{Li}_2\,y-\operatorname{Li}_2\,x-\log(1-x)\log(1-y)$$

Set $\;u:=\dfrac x{1-y},\;v:=\dfrac y{1-x}\,$ to conclude :

$$\operatorname{Li}_2(u v)=\operatorname{Li}_2(u) + \operatorname{Li}_2(v) - \operatorname{Li}_2(x) - \operatorname{Li}_2(y)-\log(1-x) \log(1-y) $$

I don't understand why do you have troubles when differentiating. Using that $$1-u=\frac{1-x-y}{1-y},\qquad 1-v=\frac{1-x-y}{1-x},\qquad 1-uv=\frac{1-x-y}{(1-x)(1-y)}$$ and that $\displaystyle\mathrm{Li}_2'(x)=-\frac{\ln(1-x)}{x}$, the derivative of the right hand side with respect to $x$ is \begin{align} &-\ln(1-u)\left(\ln u\right)'_x-\ln(1-v)\left(\ln v\right)'_x+\ln(1-uv)\left(\ln u+\ln v\right)'_x+\frac{\ln(1-x)}{x}=\\ &=-\frac{\ln(1-x-y)-\ln(1-y)}{x}-\frac{\ln(1-x-y)-\ln(1-x)}{1-x}+\\ &+\Bigl(\ln(1-x-y)-\ln(1-x)-\ln(1-y)\Bigr)\left(\frac{1}{x}+\frac{1}{1-x}\right)+\frac{\ln(1-x)}{x}. \end{align} In the last expression, the terms containing $\ln(1-x-y)$ and $\ln(1-x)$ obviously cancel out so that it reduces to $$-\frac{\ln(1-y)}{1-x}=\frac{d}{dx}\ln(1-x)\ln(1-y).$$ Therefore, to show the identity, it now suffices to verify it for some value of $x$. For example, for $x=0$ we have $u=0$, $v=y$ and the right hand side becomes $$\mathrm{Li}_2(0)+\mathrm{Li}_2(y)-\mathrm{Li}_2(0)-\mathrm{Li}_2(0)-\mathrm{Li}_2(y) =-\mathrm{Li}_2(0)=0.$$