Suppose $A$ contains at least three elements, say $a < b < c$, so $b - a$ is positive, thus it must hold that $a + c \leq b - a$. This gives $c \leq b - 2a$. Therefore, $A$ contains at most two elements.
Suppose $B$ contains at least four elements, say $a < b < c < d$. Then $A$ contains the three distinct elements $\frac{d}{a}, \frac{d}{b}$, and $\frac{d}{c}$. But $A$ cannot contain three distinct elements, contradiction. Therefore, $B$ contains at most three elements.
In total, $A \cup B$ contains at most 5 elements. This is possible, for example with $A = \{2, 4\}$ and $B = \{3, 6, 12\}$. Now $2 + 4 = 6 \in B$ and $\frac{12}{6} = \frac{6}{3} = 2 \in A$ and $\frac{12}{3} = 4 \in A$, so this pair of sets satisfies the conditions. We conclude that $A \cup B$ contains at most 5 elements.
Finding a pair of sets that satisfies the conditions can be done as follows. Suppose $B$ contains the elements $a < b < c$. Then $A$ contains the elements $\frac{c}{b}, \frac{b}{a}$, and $\frac{c}{a}$, with $\frac{c}{a}$ being the largest. Since $A$ contains only two elements, it must hold that $\frac{c}{b} = \frac{b}{a}$. Furthermore, the sum of the two elements in $A$ must be in $B$, so $\frac{b}{a} + \frac{c}{a} \in B$. Write $b = ta$, then $c = tb = t^2a$ and $\frac{b}{a} + \frac{c}{a} = t + t^2$. Now you see that $b$ can equal $t + t^2$ by choosing $a = t + 1$. With $t = 1$, you get $b = a$, so this does not work; with $t = 2$, you get the solution mentioned above.