For $x\in(-\tfrac1{\mathrm{e}},0)$ and $\alpha\in(0,1)$, defining $$\large x=\alpha ^{\frac{\alpha }{1-\alpha }} \log \left(\alpha ^{\frac{\alpha }{1-\alpha }}\right)$$ the real branches of Lambert function can be parametrized as
$$W_0(x)=\frac{\alpha \log (\alpha )}{1-\alpha }\qquad \text{and} \qquad W_{-1}(x)=\frac{ \log (\alpha )}{1-\alpha }$$
(have a look here).
This allows to find interesting inverses (see here and here) $$y=W_0(x)-W_{-1}(x) \qquad \implies \qquad \large\color{blue}{x=\frac{y}{1-e^y} e^{\frac{y}{1-e^y}}}$$
$$y=\frac{W_{-1}(x)}{W_0(x)} \qquad \implies \qquad \large \color{blue}{ x=y^{\frac{1}{1-y}}\,\frac{\log(y)}{y-1}}$$
What could we do for $$y=W_0(x)+W_{-1}(x)=\frac{1+\alpha }{1-\alpha }\,\log (\alpha )\tag 1$$ and/or for $$y=W_0(x)\times W_{-1}(x)=\frac{\alpha }{(1-\alpha )^2}\,\log ^2(\alpha )\tag 2$$ even in terms of approximations.
For $(1)$, with $\alpha=e^t$, we have $$e^{-t}=-\frac {t+y}{t-y}$$ which has a solution in terms of the generalized Lambert function.
For $(2)$, for a numerical solution $$y \sim 1.1 (-e x)^{\frac 3{4}}-0.1 (-e x)^{\frac 3{2}}$$ gives a reasonable guess.
