Show that $S^{-1}A = B$ for integral domains and $S = \{x\in A\setminus\{0\}: x^{-1}\in B\}$ Let $A \subset B$ be commutative integral domains with $\operatorname{Quot}(A) =  \operatorname{Quot}(B).$

Now consider the multiplicatively closed subset $S = \{x\in A\setminus\{0\}: x^{-1}\in B\}$. 

I want to show that $S^{-1}A = B$. I would appreciate any help.
It suffices to prove $B\subseteq S^{-1}A$, since we already have $A\subseteq B\subseteq \mathrm{Frac}(A)$.
For any $x\in B$, $x=\frac{a}{b}$ for some $a,b\in A$. By definition, correct.
It suffices to prove $B\subseteq S^{-1}A$, since we already have $A\subseteq B\subseteq \mathrm{Frac}(A)$.
For any $x\in B$, $x=\frac{a}{b}$ for some $a,b\in A$. By definition, correct.