I am looking at the following equation which is solvable in terms of the Lambert-W function when $a=0$ (but it is strictly positive in my case, i.e. $a>0$):
$x(x+a)e^x=b$ $(a,b>0)$
more generally, one can consider a generalization to the lambert function of the form : $(x-r_1)(x-r_2)e^x=b$ (in my case $r_1=0, r_2=-a$). Can the solution $x$ be expressed using the Lambert function or other special functions?
There is no closed form in terms of the Lambert W function. What you have to be careful about when dealing with the solving using the Lambert W function is addition and exponents.
It is possible to solve $(x+a)e^x=b$ but $x(x+a)e^x=b$ is not solvable.
It would require you to get the exponent to become $x(x+a)$, which means you'd have to exponentiate everything by $x+a$ to get it into the exponent. And, that leads to problems.
If you really need a solution, numerical methods or something a bit simpler like Fixed-point conversion might do the job.