Edit: easy, but incorrect. It needs to be modified to : if there is a finite head, then there must be a finite (possibly trivial) factor with empty head.
An easy observation: such a structure with a finite nonempty head is not directly decomposable.
Let sigma not be onto. That part of the base set outside the range of sigma I call the head. Then invertibility implies the base set is infinite. If one has two structures with one having a nonempty head, their product will have a nonempty head that is infinite. Therefore any such structure with a finite nonempty head is not directly decomposable. The free finitely generated structures in this variety gives a class of such examples.
Gerhard "Nothing Up My Sleeve... Presto!" Paseman, 2020.05.19.