I've tried to solve $x\log x = 2(x-1)(1-ax)$ for $a\ge 0$. If $a=0$, I obtained $$x \log x = 2x-2\\ \to x(\log x-2)= -2 \\ \to x\log(xe^{-2}) = -2 \\ \to xe^{-2}\log(xe^{-2}) = -2e^{-2}\\ \to x =e^{W(-2e^{-2})+2}.$$
Similarly as in the above solution, can I express the solution for $a>0$ in terms of Lambert $W$ function?
Of course $x=1$ is a solution, but that's presumably not the one you want.
As far as I can tell, there is no closed-form expression for the solution when $a > 0$, whether in terms of $W$ or anything else. As a series in powers of $a$: with $w = W(-2 e^{-2})$,
$$ x = e^{2+w} \left( 1+2\,{\frac {w+2}{w \left( w+1 \right) }}a+2\,{\frac { \left( w+2 \right) \left( 2\,{w}^{2}+9\,w+6 \right) }{{w}^{2} \left( w+1 \right) ^{3}}}{a}^{2}+\ldots\right)$$
It appears that the coefficient of $a^n$ in the series is $(w+2)/(w^n (w+1)^{2n-1})$ times a polynomial in $w$ of degree $2n-2$.
