If I understand the definition precisely, $S = 3\mathbb{Z} \cup \{2\}$ works.
$S$ is not an additive basis, as any sum of elements in $S$ is either a multiple of 3 or 2 more than a multiple of 3.
But, since $S^2$ also contains 4, we can now write all integers as a sum of at most 3 elements from $S \cup S^2$.
If you are allowed to 're use' elements, this example doesn't work, but perhaps is useful nonetheless?