Background. This question belongs to evil mathematics. It is motivated by this question which links to a paper in which it is claimed that it is an open problem whether there exists strictly functorial choices for pullbacks in the category of sets. While thinking about this problem, I had the idea to warm up with coproducts:
Definition. Let $C$ be a category with coproducts. Then for all objects $X,Y,Z$ we may choose the iterated coproducts $X \coprod (Y \coprod Z)$ and $(X \coprod Y) \coprod Z$. If we have choosen them, there is a canonical isomorphism $\alpha$ between these coproducts. Let's call $X,Y,Z$ strictly associative (with respect to coproducts) if the coproducts may be chosen in such a way that this isomorphism equals the identity. This means, in particular, that the objects $X \coprod (Y \coprod Z) = (X \coprod Y) \coprod Z$ are equal.
Question. In the category (Set) of sets and maps, is it true that every triple of objects is strictly associative? In other words (without category theoretic terms), is it possible to choose for every pair of sets $X,Y$ a set $X \coprod Y$ which is the union of two disjoint sets equipped with bijections to $X$ and $Y$, such that $X \coprod (Y \coprod Z) = (X \coprod Y) \coprod Z$ as a disjoint union of three sets equipped with bijections to $X$, $Y$ and $Z$?