I don't know how satisfactory this will be, but at least its a first stab at an answer, and might highlight some of the issues.
There is one "obvious" condition which ensures $f^\times$ is surjective: if the kernel of $f$ is contained in the Jacobson radical of $R$, then $f^\times$ is surjective. We can think of $S$ as being $R/I$ for some ideal $I$, so that maximal ideals of $R/I$ correspond to maximal ideals of $R$ containing $I$. Since units are precisely elements that miss all maximal ideals, if every maximal ideal of $R$ contains $I$ then every unit in $R/I$ can be lifted to a unit in $R$ (in fact, every lift to an element of $R$ is a unit in this case).
For $I$ not contained in the Jacobson radical, $R$ will have maximal ideals not containing $I$, and the question of whether every unit in $R/I$ lifts to an element of $R$ missing every maximal ideal in $R$ seems subtle.
There are probably other, better, weaker conditions which will imply surjectivity, however.
It is also useful to keep in mind the following example: the map $k[x] \to k[x]/(x^2)$ is surjective and does not induce a surjection on units.