There is a natural way to extend a binary operation $\star$ on $\mathbb{Z}$ to a semicontinuous binary operation $\widehat{\star}$ on $\beta\mathbb{Z}$, the latter being the set of ultrafilters on $\mathbb{Z}$; see here (replacing $\mathbb{N}$ with $\mathbb{Z}$ as appropriate). In general, algebraic properties of $\star$ are not preserved by this extension - e.g. $\widehat{+}$ has nontrivial idempotents and is non-commutative - but this extension can nonetheless be useful.
I'm interested in understanding how noncommutative $\widehat{+}$ is (right now, the best result I know is that it has trivial center). Normally one might attack this sort of question by looking at the commutator subgroup of $(\beta\mathbb{Z};\widehat{+})$, but since $\widehat{+}$ is non-cancellative this doesn't make sense. We can however whip up a "brute force" analogue by considering $\widehat{+}$ and $\widehat{-}$ explicitly. This leads to the following:
Question 1: Is every nonprincipal ultrafilter $\mathcal{U}$ on $\mathbb{Z}$ equal to $$[(\mathcal{W}_1\widehat{+}\mathcal{W}_2)\widehat{-}\mathcal{W}_1]\widehat{-}\mathcal{W}_2$$ for some ultrafilters $\mathcal{W}_1,\mathcal{W}_2$ on $\mathbb{Z}$?
Of course, this is a rather naive attempt to whip up a commutator-like notion:
Question 2: Is there a more useful analogue of the notion of commutator when looking at non-cancellative semigroups (namely $(\beta\mathbb{Z};\widehat{+})$)?
I know that there is a lot of universal algebra around generalized commutator theory, but to my limited understanding it works best when the algebra in question is (say) congruence modular, and it's not clear to me whether $(\beta\mathbb{Z};\widehat{+})$ is congruence modular.