If $R$ is a local ring and $S$ is its residue field then the map is onto, but that's too easy, isn't it?
I don't think this is a trivial question at all! For example, consider the ring ${\mathbf Z}[\sqrt{2}]$, which has infinitely many units ($\pm 1$ times powers of $1+\sqrt{2}$). For any nonzero prime ideal $(\pi)$ (the ring is a PID so the ideal is principal, not that it matters), we can reduce mod $\pi$ and get a map ${\mathbf Z}[\sqrt{2}] \rightarrow {\mathbf Z}[\sqrt{2}]/(\pi)$. This is onto and the target ring is a finite field, so its unit group is cyclic.
Asking whether the map of unit groups is onto is essentially equivalent to asking
if $1 + \sqrt{2}$ is a generator of the units mod $\pi$. This doesn't always happen (e.g., when $\pi = 5$ the ring ${\mathbf Z}[\sqrt{2}]/(5)$ is a field of size 25,
$1+\sqrt{2} \bmod 5$ has order 12, and $(1+\sqrt{2})^{6} \equiv -1 \bmod 5$, so
the whole unit group of ${\mathbf Z}[\sqrt{2}]$ maps onto only half the units mod 5).
However, it is conjectured that there are infinitely many prime ideals $(\pi)$ such that $1+\sqrt{2} \bmod \pi$ is a generator of the units. This is still an open problem, although it is known to follow from suitable versions of the Generalized Riemann Hypothesis.
This is a generalization of Artin's primitive root conjecture, which says that any nonzero integer $a$ other than $\pm 1$ or a perfect square should be a generator of the units mod $p$ for infinitely many primes $p$. For example, $10 \bmod p$ should be a generator for infinitely many $p$. (Concretely, this says there should be infinitely many $p$ such that $1/p$ has decimal period $p-1$, which is the longest it could conceivably be for any $p$.) Artin's original conjecture may not seem like it fits your specific question, since ${\mathbf Z}$ has only two units, but it is straightforward to make Artin's problem
fit your question, e.g., use ${\mathbf Z}[1/10]$ instead of ${\mathbf Z}$ and its unit group is $\pm 10^{\mathbf Z}$. Artin's conjecture for $a=10$ amounts to saying the unit group of ${\mathbf Z}[1/10]$ maps onto the unit group of its reduction modulo infinitely many primes (not counting 2 and 5, which are no longer prime).
