I do not think that you could get a closed form for $x$ and numerical methods will be required.
To isolate the important terms, let
$$k=\frac{\lambda}{\ln{\frac{1-a}{a}}}-a \qquad \text{and} \qquad b=(k+1)\log(1-a)-k\log(a) $$ which make that we look for the non-trivial zero $(x=a)$ of the function
$$f(x)=b+k\log(x)-(k+1)\log(1-x)$$
When $\lambda$ is "close" to its lower bound, $x$ is "small". Expanding $f(x)$ around $x=0$ gives
$$f(x)=b+k \log (x)+(k+1) x+O\left(x^2\right)$$ Ignoring the higher order terms, the "approximate" solution is
$$x=\frac k {k+1}\,W\left(\frac{k+1}{k}e^{-\frac{b}{k}}\right)$$ where appears Lambert function.
Similarly, when $\lambda$ is "close" to its upper bound, $x$ is "close" to $1$. Expanding $f(x)$ around $x=1$ gives
$$f(x)=b-k (1-x)-(k+1) \log (1-x)+O\left((1-x)^2\right)$$Ignoring the higher order terms, the "approximate" solution is
$$x=1-\frac {k+1}k\,W\left(\frac k{k+1}e^{-\frac{b}{k+1}}\right)$$
When $\lambda$ is "close" to $0$, $x \sim \lambda$.
For example, with $a=0.7$ and $\lambda=-0.3$, the approximation gives $x=0.0843$ while the exact solution, obtained using Newton method, is $0.0850$.
Edit
From a practical point of view, for a given $a$, I should build a table of $\lambda$ as a function of $x$ (this is a direct calculation) and, later, for a given $\lambda$, interpolate in the table to obtain a starting guess $x_0$ for starting Newton iteration.
Using again $a=0.7$, the table would be
$$\left(
\begin{array}{cc}
x & \lambda \\
0.01 & -0.407234 \\
0.02 & -0.381465 \\
0.03 & -0.363124 \\
0.04 & -0.348276 \\
0.05 & -0.335532 \\
0.10 & -0.287362 \\
0.15 & -0.251336 \\
0.20 & -0.221038 \\
0.25 & -0.194133 \\
0.30 & -0.169460 \\
0.35 & -0.146334 \\
0.40 & -0.124303 \\
0.45 & -0.103039 \\
0.50 & -0.082283 \\
0.55 & -0.061814 \\
0.60 & -0.041424 \\
0.65 & -0.020900 \\
0.70 & +0.254189 \\
0.75 & +0.021582 \\
0.80 & +0.044279 \\
0.85 & +0.068787 \\
0.90 & +0.096449 \\
0.95 & +0.130807 \\
0.96 & +0.139368 \\
0.97 & +0.149046 \\
0.98 & +0.160548 \\
0.99 & +0.175825
\end{array}
\right)$$