After asking this question and getting an answer (a counterexample), I wish to further assume that the ring extension $A \subseteq B$ is separable, namely, $B$ is a projective $B \otimes_A B$-module, see also this blog.
More elaborately, Let $A \subseteq B$ be an algebraic separable ring extension of integral domains having fields of fractions $Q(A) \subseteq Q(B)$. Further assume that $Q(A) \cap B = A$. Take $b \in B-A$, $a_1,a_2,a_3,a_4 \in A$ with $a_2a_4 \neq 0$, and denote $w:=\frac{a_1}{a_2}+\frac{a_3}{a_4}b$. Of course, $w \in Q(B)$.
Is it true that if $w \in B$, then $\frac{a_1}{a_2} \in A$ and $\frac{a_3}{a_4}\in A$?
Notice that it is enough to show that $\frac{a_1}{a_2} \in B$ and $\frac{a_3}{a_4}\in B$, since then $Q(A) \cap B = A$ implies that $\frac{a_1}{a_2} \in A$ and $\frac{a_3}{a_4}\in A$.
Perhaps it would be easier to answer my question if we further assume that: (1) $A^*=B^*=k^*=k-\{0\}$, where $k$ is the base field; in that case, my question becomes: Is it true that if $w \in B$ then $a_2 \in k^*$ and $a_4 \in k^*$? (2) Both $A$ and $B$ are UFD's.
Thank you very much for any help.