Here is a similar example, with the minority operator: the minority of three languages $A,B,C$ consists of words belonging to at most 1 of $A,B,C$.
If $A,B,C$ are decidable then so is their minority. Indeed, given a word $w$, first determine whether $w$ belongs to $A$, to $B$, and to $C$. You can do this since $A,B,C$ are decidable. Depending on this data, either accept or reject $w$. This is an algorithm deciding the minority.
If $A,B,C$ are enumerable, then their minority need not be so. Indeed, let $A$ be a language which is enumerable but not co-enumerable (for example, the set of halting programs). The minority of $A,A,A$ consists of all words not in $A$, i.e., the complement of $A$. By assumption $A$ is not co-enumerable, so the minority of $A,A,A$ is not enumerable.