I think it relates to a Gumbel distribution:
if I define $S=R^2$, when $R\sim p(r|\lambda)$, the density of $S$ is given by the Jacobian formula:
\begin{align*}
q(s|\lambda) &= p(\sqrt{s}|\lambda)\times \left|\frac{\text{d}r}{\text{d}s}\right|\\
&= p(\sqrt{s}|\lambda)\times \frac{1}{2\sqrt{s}}\\
&= \frac{2\lambda \sqrt{s}\exp\left(\lambda\exp\left(-s\right)-s\right)}{\exp\left(\lambda\right)-1}\,\frac{1}{2\sqrt{s}}\\
&= \frac{\lambda \exp\left(\lambda\exp\left(-s\right)-s\right)}{\exp\left(\lambda\right)-1}\\
&= \frac{\exp\left(\exp\left(\log\{\lambda\}-s\right)+\log\{\lambda\}-s\right)}{\exp\left(\lambda\right)-1}\\
&= \frac{\exp\left(\exp\left(-z\right)-z\right)}{\exp\left(\lambda\right)-1}\\
\end{align*}
where $z=s-\log\{\lambda\}$
So $S$ is almost distributed as a Gumbel distribution with parameter $(\log\{\lambda\},1)$ except that (a) its support is truncated to $(0,+\infty)$ and (b) there is a missing - in front of the exponential inside the exponential...
This means that $S-\log\{\lambda\}$ has a fixed distribution with the above density and cdf $F$. From this representation, there exists a transform $G^{-1}\circ F$ (with $G$ being the cdf of the Gumbel distribution) that turns $ S-\log\{\lambda\}$ into a standard Gumbel, but this is not very useful!
Note that, in the mixture representation, $S=R^2$ is then distributed as an Exponential $\mathcal{E}(n)$ variate.