If $P(A \ \cup \ B) = P(A) + P(B)$, is it the case that $A$ and $B$ are disjoint? I know that if $A$ and $B$ are disjoint events, then $P(A \cup \ B) = P(A) + P(B)$. However, is the converse true as well? Thanks.
No, all you can deduce is that $P(A \cap B) = 0$, because $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ It doesn't mean that $A \cap B$ is empty.
EDIT: This is incorrect, disregard. The comments have valuable discussion and counterexamples.

Yes. Assuming $A$ and $B$ are independent, by definition, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. If we have $P(A \cup B) = P(A) + P(B)$, then $P(A \cap B) = 0$, which is the definition of disjoint events.