Birational and faithfully flat $\implies$ isomorphism

We can show $I = A$ as follows, the inclusion $I \subseteq A$ being clear. Choose $a \in A$ and write $b = a_1/a_2$. Then $aba_2 = a\cdot a_1 \in A$ and so $a\cdot a_1 \in a a_2 B \cap A = aa_2 A$ where the last equality comes from faithful flatness. It follows that $aba_2 \in aa_2A$ and so $aba_2 = a a_2 a'$ for some $a' \in A$. Thus $ab = a a' \in A$ and so indeed $a \in I$. Thus $I = A$.

As pointed out in the comments, the problem is easy if we know that $B$ is a localisation of $A$, because then the codiagonal $\nabla : B \otimes_A B \to B$ is an isomorphism, and so we can (co)base change $A \to B$ along itself to deduce that $A \to B$ is an isomorphism. (This is where we use the fact that $A \to B$ is faithfully flat.) But in fact $\nabla : B \otimes_A B \to B$ is an isomorphism if and only if $A \to B$ is an epimorphism, and this is certainly true if $B$ is flat and embeds in $\operatorname{Frac} A$ as an $A$-algebra. (See comments below.)