TL;DR: Yes, in a finite dimensional Hilbert space, $G^{-1}$ can be approximated to arbitrary accuracy as a finite power of $G$. If $G$ has finite order $r$, this follows from $G^{-1}=G^{r-1}$. The general case can be proven using the simultaneous version of Dirichlet's approximation theorem. Consequently, for any set of gates $C$, if $C\cup\{G,G^{-1}\}$ is universal$^1$, then so is $C\cup\{G\}$.
It is sufficient to show that the set $\{G^n\,|\,n=1,2,\dots\}$ contains a sequence $G_i$ with $i=1,2,\dots$ such that $\lim_{i\to\infty}G_i=I$.
Suppose that $G=\sum_{j=1}^d e^{2\pi i\alpha_j}|\psi_j\rangle\langle\psi_j|$ for some $\alpha_j\in[0,1)$ and some orthonormal basis $|\psi_j\rangle$. The simultaneous version of Dirichlet's approximation theorem asserts that for every positive integer $k=1,2,\dots$, there are integers $p_{k,1},\dots,p_{k,d}$ and $q_k$ such that $1\leqslant q_k\leqslant k$ and $\Delta_{k,j}:=q_k\alpha_j-p_{k,j}\in\left[-\frac{1}{k^{1/d}},\frac{1}{k^{1/d}}\right]$. Consequently, $\lim_{k\to\infty}\Delta_{k,j}=0$ for every $j$.
Define $G_k:=G^{q_k}$. I will use entanglement fidelity $F_e$ as a measure of closeness between $G_k$ and $I$. For a general quantum channel $\mathcal{E}$, it is defined as
\begin{align}
F_e(\mathcal{E}):=\langle\psi|(\mathcal{E}\otimes\mathcal{I})(|\psi\rangle\langle\psi|)|\psi\rangle\tag1
\end{align}
where $\mathcal{I}$ is the identity channel and $|\psi\rangle:=\frac{1}{\sqrt{d}}\sum_k|k\rangle|k\rangle$ is the a maximally entangled state. For a unitary channel $\mathcal{U}(\rho)=U\rho U^\dagger$ we have $F_e(U):=F_e(\mathcal{U})=\frac{1}{d^2}|\mathrm{tr}(U)|^2$, so
\begin{align}
\lim_{k\to\infty}F_e(G_k)&=\frac{1}{d^2}\lim_{k\to\infty}|\mathrm{tr}(G^{q_k})|^2\tag2\\
&=\frac{1}{d^2}\lim_{k\to\infty}\left|\sum_j e^{2\pi iq_k\alpha_j}\right|^2\tag3\\
&=\frac{1}{d^2}\lim_{k\to\infty}\left|\sum_j e^{2\pi i\Delta_{k,j}}\right|^2\tag4\\
&=\frac{1}{d^2}d^2=1\tag5
\end{align}
where the penultimate equality follows from the continuity of the exponential function and the fact that $\lim_{k\to\infty}\Delta_{k,j}=0$ for every $j$. Finally, unitary group is compact, so $G_k$ has a convergent subsequence $G_i$. Let $G':=\lim_{i\to\infty} G_i$. Entanglement fidelity $F_e(U)$ is a continuous function of $U$, so by calculation above $F_e(G')=1$. But this implies that $G'=I$, so $\lim_{i\to\infty}G_i=I$.
$^1$ I assume that universal means "generates a dense subset of the unitary group". A weaker, but perhaps more important, concept of universality is the computational universality which means "able to realize arbitrary quantum computations". An example of a computationally universal set of unitaries that fails to be universal is the Hadamard and Toffoli.
