Inverse of $f(x)=x/\log(1+x)$ for $x>1$

Question

Is there an expression of $f^{-1}$ in terms of product-log or other special functions which holds for $x>1$?

WolframAlpha finds the solution: $$g(y)=-y\,W\big(-\mathrm e^{-1/y}/y\big)-1,$$ which however only holds for $-1<x<1$, $0<y<1$.

Edit: Corrected $g(y)$ as per @jjagmath's comment.

Answer 1

As user1337 said in his comment, you just need to consider the other branch of the Lambert function: