4 correction

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 $b = u - v$.

I think the answer is no, but I didn't get anywhere. Here is a partial negative answerthe condensed version of what I posted earlier, maybe somebody else can finish it off...which seems to put serious constraints on $S$.
Let $R$ be any field . Suppose that and let $S$ is a proper be an additive subgroup of $R$ which happens to be is closed under multiplication. Pick $r \in R \setminus S$. Let $x,y \in S$ be distinct and consider the difference $d = (r-x)^{-1} - (r-y)^{-1}$. If $d \in S$ then $r$ satisfies the quadratic equation $dr^2-d(x+y)r+(dxy-y+x) = 0$ with coefficients in $S$. So if $r$ does not satisfy a quadratic equation with coefficients in $S$, then the reals $(r-x)^{-1}$ for $x \in S$ must all belong to different additive cosets of $S$ in $R$. Hence the index of $S$ is at least $|S|$.
Actually, if $S$ has index less than $|S|$, |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$. 2 addendum I think the answer is no. Here is a partial negative answer, maybe somebody else can finish it off... Let$R$be any field. Suppose that$S$is a proper additive subgroup of$R$which happens to be closed under multiplication. Pick$r \in R \setminus S$. Let$x,y \in S$be distinct and consider the difference$d = (r-x)^{-1} - (r-y)^{-1}$. If$d \in S$then$r$satisfies the quadratic equation$dr^2-d(x+y)r+(dxy-y+x) = 0$with coefficients in$S$. So if$r$does not satisfy a quadratic equation with coefficients in$S$, then the reals$(r-x)^{-1}$for$x \in S$must all belong to different additive cosets of$S$in$R$. Hence the index of$S$is at least$|S|$. Actually, if$S$has index less than$|S|$, then every element of$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\$.