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No. The ten residue classes modulo $10$ form a finite partition, so an ultrafilter $p$ contains exactly one class $C_r$. The addition formula shows that $p+p$ contains $C_{2r}$: for $n\in C_r$, the translate $C_{2r}-n$ is $C_r$ and belongs to $p$. If $p$ is an idempotent ultrafilter, uniqueness of its selected class gives $2r\equiv r(\mathrm{mod}10)$, hence $r=0$.

Thus every idempotent ultrafilter contains the multiples of $10$, and cannot also contain their complement. This is the zero-residue constraint for idempotent ultrafilters, valid for every finite modulus.