I think the answer is no, but I didn't get anywhere. [Edit: I used to think the answer was no, but Simon Thomas convinced me otherwise.] Here is the condensed version of what I posted earlier, which seems to put serious constraints on $S$.
Let $R$ be any field and let $S$ be an additive subgroup of $R$ which is closed under multiplication. If $S$ has index less than $|S|$ as an additive subgroup of $R$, then every element of $R$ is of the form $a/b$ for $a, b \in S$. To see this, pick $r \in R$ and consider the multiples $ur$ for $u \in S$. Since $S$ has index less than $|S|$, there must be $u \neq v$ such that $a = ur - vr \in S$ then $r = a/b$ where $b = u - v$.