At best Übermensch wants to know if expanding the search for solutions to matrices might improve the situation. It does not.
Assuming that matrices with entries in $\mathbb{C}$ are intended, the RHS series will diverge except on nilpotent matrices. For any nilpotent $x$, since the RHS terms all contain at least one factor $x$, the series can never converge to a non-singular limit. In particular the identity cannot be the "limit" (which I put in quotes since for nilpotent $x$, only finitely many terms of the series are nonzero).
Added: The comments I made on another Answer posted here show that no solution is possible if convergence is taken as in formal power series. Any mapping of $\mathbb{Z}[[X]]$ to ring R such that the image $x$ of $X$ satisfies the equation $1 = \sum_{n \gt 0} n^n x^n$ would necessarily map the unit $1 - \sum_{n \gt 0} n^n X^n$ to zero, and thus the kernel would be all of $\mathbb{Z}[[X]]$.