Let $A \in \mathbb{R}^{n \times n}$ be any matrix. Is it true that if $m \in \mathbb{R}$ is such that $|m|\cdot||A||<1$, then $$A^{-1}=\sum_{n=0}^{\infty}(-1)^nm^{n+1}(A-\frac{1}{m}I)^n\,,$$ or is it false? I just substituted $A=B+I \frac{1}{m}$ in the original Neumann series saying that
$$(B+I\frac{1}{m})^{-1}=\sum_{n=0}^{\infty}(-1)^nm^{n+1}B^n\,.$$
If $S=\sum_{n=0}^{\infty}(-1)^nm^{n+1}\left(A-\frac{1}{m}I\right)^n$ ever converges, it is straightforward to verify that $AS=I$ and hence $S=A^{-1}$. The only question is whether the infinite series converges or not. Since this can be rewritten as a Neumann series in $B$, we know that it doesn't always converge. Convergence is guaranteed when $\|mB\|=\|mA-I\|<1$, but the series may diverge otherwise. For instance, let $A=-\epsilon I$ for some $\epsilon>0$, so that $A^{-1}$ always exists. However, when $\epsilon$ is small and $m=1$, we have $\|mA\|=\epsilon\|I\|<1$ but $$ \sum_{n=0}^{\infty}(-1)^nm^{n+1}\left(A-\frac{1}{m}I\right)^n =\sum_{n=0}^\infty (1+\epsilon)^nI $$ doesn't converge.