Show that a set is a ring

Question

Let $R\not=\{0\}$ be a commutative ring with unity. Let $I$ be a prime ideal in $R$. Let $S=R-I=\{x\in R|x\not\in I\}$. Let $F$ be a field that contains $R$ as a subring with the same unity. Show that $A=\{ab^{-1}\in F|a\in R, b\in S\}$ is a ring. I can easily show that $(A, \cdot)$ is a semigroup but I can't find a way to prove that $(A, +)$ is has a closure to addition. I've tried showing that if $ab^{-1},cd^{-1}\in A$ than there is a product of $(a+c+\dots)(b^{-1}+d^{-1}+\dots)$ but it did not work. I would like a hint for that.

Answer 1

I explain Mark Bennet's comment :
if $ab^{-1},cd^{-1}\in A$ than $ab^{-1}+cd^{-1}=(ad+bc)(bd)^{-1}$. since $I$ is prime $bd\in S$.