Localization - the equivalence ratio and subrings.

Let $R$ be a commutative ring with $1\in R$, and let $S\subseteq R$ a multiplicatively closed set. Let $A\subset R$ be a subring (with $1\in A$), and observe $S^{-1}A\subseteq S^{-1}R$.

Is it possible that there exists $t\in R, s\in S$, such that $\frac{t}{s}\in A^{-1}S$ and $t \in R\setminus A$?

We need to prove that there exist $s_1,s^*\in S, a\in A $ such that $\frac{t}{s}=\frac{a}{s^*}$, that is $s_1(s^*t-sa)=0$, but I have no idea how to start (if it is indeed true).

If $S^{-1}A\triangleleft S^{-1}R$ is an ideal ($1\notin A$, we could reduce this question to $\frac{t}{1}\in A^{-1}S$ by multiplication with $\frac{s}{1}$ and $t\in R\setminus A$.

Back to the question, is it possible that $\frac{t}{s}\in A^{-1}S$ with the above conditions?

Thanks in advance.

Consider $R=\Bbb Z[i]$, $S:=\{(1+i)^n: n\in\Bbb N\}$ and $A:=\Bbb Z$.

Then in $S^{-1}A$ we have $$\frac 2{1+i}=1-i=\frac{1-i}1$$ where the numerator is $\in R\setminus A$.