It could be interesting to use the parametrization of Lambert function (have a look here).
For $x\in(-\tfrac1{\mathrm{e}},0)$ and $\alpha\in(0,1)$, defining
$$\color{blue}{\large x=\alpha ^{\frac{\alpha }{1-\alpha }} \log \left(\alpha ^{\frac{\alpha }{1-\alpha }}\right)}$$
the real branches of Lambert function can be parametrized as
$$\color{blue}{W_0(x)=\frac{\alpha \log (\alpha )}{1-\alpha }}\qquad \text{and} \qquad \color{blue}{W_{-1}(x)=\frac{ \log (\alpha )}{1-\alpha }}$$
For the first function
$$y=\frac{W_0(x)\,(W_{-1}(x)+1)}{(W_0(x)+1)\,W_{-1}(x)}=\frac{\alpha \left(\frac{\log (\alpha )}{1-\alpha
}+1\right)}{\frac{\alpha \log (\alpha )}{1-\alpha }+1}$$ which cannot be inversed but quite well approximated
by the $[n,n]$ Padé approximant $P_n$ built around $\alpha=1$
This would give
$$P_2= -\frac{11 \alpha ^2+18 \alpha +1}{\alpha ^2+18 \alpha +11}$$
$$P_3=-\frac{57 \alpha ^3+239 \alpha ^2+121 \alpha +3}{3 \alpha ^3+121
\alpha ^2+239 \alpha +57}$$
$$P_4=-\frac{29 \alpha ^4+227 \alpha ^3+300 \alpha ^2+73 \alpha
+1}{\alpha ^4+73 \alpha ^3+300 \alpha ^2+227 \alpha +29}$$
The respective errors of these approximants are
$$\frac{(\alpha -1)^5}{525} \qquad \qquad \frac{(\alpha -1)^7}{8820} \qquad \qquad \frac{(\alpha -1)^9}{145530}$$
Considering the infinite norm
$$\Phi_n=\int_0^1 \Bigg( \frac{\alpha \left(\frac{\log (\alpha )}{1-\alpha
}+1\right)}{\frac{\alpha \log (\alpha )}{1-\alpha }+1}-P_n \Bigg)^2\, d \alpha$$ they are
$$\Phi_2=1.99614\times 10^{-4} \quad\quad \Phi_3=3.70724\times 10^{-5} \quad\quad \Phi_4=1.01236\times 10^{-5}$$
So, at the price of a quadratic, cubic or quartic equation, we can have an approximate value of $\alpha$ and then $x$ from its definition.
A few numbers (using the chain $x\to \alpha \to y \to \alpha \to x$)
$$\left(
\begin{array}{cccc}
x & x_{(2)} & x_{(3)} & x_{(4)} \\
-0.15 & -0.113812 & -0.140978 & -0.147205 \\
-0.20 & -0.185725 & -0.197052 & -0.199294 \\
-0.25 & -0.245433 & -0.249309 & -0.249885 \\
-0.30 & -0.299159 & -0.299925 & -0.299993 \\
-0.35 & -0.349985 & -0.350000 & -0.350000 \\
\end{array}
\right)$$
Using $P_3$, the cubic to be solved for $\alpha$ is
$$3 (y+19)\, \alpha ^3+ (121 y+239)\,\alpha ^2+ (239 y+121)\,\alpha+3(19y+1)=0$$
With
$$p=\frac{10 (1249 y^2+1588 y+3643}{27 (y+19)^2}$$
$$q=\frac{40 (69404 y^3+118731 y^2+222576 y+318289)}{729 (y+19)^3}$$ the approximate solution is
$$\alpha=2 \sqrt{\frac{p}{3}}\cos \left(\frac{1}{3} \cos ^{-1}\left(\frac{3 \sqrt{3} q}{2p^{\frac 32} }\right)\right)-\frac{121 y+239}{9 (y+19)}$$
Another thing which could be done is a series expansion around $\alpha=1$ followed by a power series reversion yo obtain
$$\alpha=1-t +\frac {t^2}6 \sum_{n=0}^\infty a_n\,t^n \qquad \text{where} \qquad t= \frac 32(1+y)$$ the very first coefficients being
$$\left\{1,\frac{2}{15},\frac{1}{36},\frac{71}{9450},\frac{281}{11340
0},\frac{29}{30375},\frac{16859}{40824000},\frac{36557}{18860688
0},\frac{27230983}{282910320000}\right\}$$
Using the above terms, for $\alpha=\frac 1e$, the exact value is $y= -\frac{1}{(e-2) e}$ and the recomputed value is $\alpha=\color{red}{0.367879}001$
