Is the Lambert W function multivalued everywhere except at $x=0$?
It is obvious that $W(0)=0\implies 0=0e^0$ because $e^u\ne0$, therefore, it is the coefficient that determines such, and the only coefficient that will work is $0$.
But are all other $x$ values going to produce a multivalued function?
Lambert W-function is multivalued almost everywhere. We can prove it using Great Picard's Theorem. Consider a holomorphic function $$f(z)=ze^z$$ The function $f$ and $W$ are inverses of each other. Moreover, $f$ has an essential singularity at $z=\tilde{\infty}$. It means that in every neighborhood of infinity (i.e. $|z|>r$) the function $f$ takes on almost all possible complex values infinitely often. It implies that its inverse is defined on almost all points, and on every point has infinitely many values.
To be specific, $f$ takes value $0$ only once, so $W$ is multivalued for all points except $0$.