Let $W_{k}(z)$ be the kth branch of the Lambert W function.
My question pertains to the branch point that the principal branch $W_{0}(z)$ shares with $W_{-1}(z)$ and $W_{1}(z)$ at $z = - \frac{1}{e}$.
By the inverse function theorem, $z=f(w)=we^{w}$ is not invertible in the neighborhood of a point where $f'(w) = e^{w}(1+w)=0$. So $f(w)$ is not invertible in the neighborhood of $w=-1$.
The principal branch of the Lambert W function assumes the value $-1$ at the point $z= -\frac{1}{e}$. And by choice of closure, $W_{-1}(-\frac{1}{e})=-1$. The branch $W_{1}(z)$ does not include any portion of the real axis.
So why then does $W_{1}(z)$ have a branch point at $z= -\frac{1}{e}$ when $W_{1}(-\frac{1}{e}) \ne -1$?
EDIT:
Sources that state $W_{1}(z)$ has a branch point at $z = - \frac{1}{e}$ include Wolfram Alpha, Maple, and this paper (page 17).
What also bothers me is that the singularity in question is a square-root singularity. (Notice that expanding at $w=-1$, we get $z = -\frac{1}{e} +\mathcal{O} \left((w+1)^{2}\right)$.) Three branches sharing a square-root singularity seems a bit strange.
This may be strangeness due to the definition of $\mathrm{W}_1$ and $\mathrm{W}_{-1}$ in Mathematica (which is where I assume you are getting the definitions for the $\mathrm{W}_k$).
Here is a plot for $k=-1,0,1$ of $|\mathrm{W}_k(z+1/e)|$ near $z=-\frac1e$:
$\mathrm{W}_0$ is in red on the bottom sheet, but $\mathrm{W}_1$ (in blue) and $\mathrm{W}_{-1}$ (in green) are split between two sheets. Only one sheet seems to share the branch point with $\mathrm{W}_0$, but both $\mathrm{W}_1$ and $\mathrm{W}_{-1}$ are part of that sheet in each neighborhood of the branch point.
Things look a little nicer if we plot the functions $|\mathrm{W}_{\text{green}}(z+1/e)|$ and $|\mathrm{W}_{\text{blue}}(z+1/e)|$ where $$ \mathrm{W}_{\text{green}}(z)=\left\{\begin{array}{} \mathrm{W}_{-1}(z)&\text{if }\mathrm{Im}(z)\ge0\\ \mathrm{W}_1(z)&\text{if }\mathrm{Im}(z)\lt0\\ \end{array}\right. $$ $$ \mathrm{W}_{\text{blue}}(z)=\left\{\begin{array}{} \mathrm{W}_1(z)&\text{if }\mathrm{Im}(z)\ge0\\ \mathrm{W}_{-1}(z)&\text{if }\mathrm{Im}(z)\lt0\\ \end{array}\right. $$
Now, only $\mathrm{W}_{\text{green}}$ shares the branch point with $\mathrm{W}_0$.
Why the Definition is as Given
I started wondering why the definition was as given. I noticed that in the plots I had made, there was no interaction between $\mathrm{W}_{\text{green}}$ and $\mathrm{W}_{\text{blue}}$. I considered increasing the domain to include the origin. Here is $|\mathrm{W}(z+1/e)|$ over a larger domain, still centered on $-\frac1e$:
This looks pretty much the same as before, when looking along the negative real axis. However, we see the reason for defining the branches as is done in Mathematica, Maple, and the paper cited in the question if we look along the positive real axis:
Viewed from this direction, we see that both the blue sheet ($\mathrm{W}_1$) and the green sheet ($\mathrm{W}_{-1}$) pass smoothly through the positive real axis.
However, as can be seen in the diagram above this one, this means that both the blue and green sheet touch the branch point on different sides; $\mathrm{W}_1$ (blue) on the negative imaginary side and $\mathrm{W}_{-1}$ (green) on the positive imaginary side.
Summary
$\mathrm{W}_0$, $\mathrm{W}_1$, and $\mathrm{W}_{-1}$ are defined to be smooth near the positive real axis. Along the negative real axis, the branch cut, the sheets are smooth if we look at the transitions. As we travel counterclockwise around the branch point, $-\frac1e$, we see the green sheet ($\mathrm{W}_{-1}$) descend, passing through the blue sheet ($\mathrm{W}_1$), then passing through the branch cut to become the red sheet ($\mathrm{W}_0$). The red sheet ($\mathrm{W}_0$) then ascends to become the blue sheet ($\mathrm{W}_1$) after passing through the branch cut. The blue sheet ($\mathrm{W}_1$) then passes through the green sheet along the real axis. Unfortunately, the branch point, $-\frac1e$, lies along the branch cut. Thus, the sheets for both $\mathrm{W}_1$ (on the negative imaginary side) and $\mathrm{W}_{-1}$ (on the positive imaginary side) touch $-\frac1e$.