Note that by this I mean Z[[x]], a commutative ring with 1 and integer coefficients.
I've been racking my mind over this question. Considering the neutral multiplicative power here is 1 and thus is an inverse of itself, wouldn't all elements of the set have a multiplicative inverse of x^-i, i being the power of Z[[x]]? Or considering there are no negative powers in the original R[[x]], would there be no elements with inverses?
If you have an invertible power series $\sum a_ix^i$ in $R[[X]]$, with inverse $\sum b_jx^j$, then $a_0b_0 = 1$, so $a_0$ has to be invertible in $R$.
Conversely, if $a_0$ is invertible in $R$, you can divide by its inverse to reduce the problem to the case of a series starting with $1$, and then apply the power series $(1+z)^{-1} = 1 - z + z^2 - z^3 + z^4 \ldots$ with $z = \sum_{i \ge 1} a_ix^i$ to find its inverse.