Deep-cut puzzles would be good examples of puzzles where commutators might not be the best way to solve, I think. The general strategy of using commutators (and conjugated commutators) works for many puzzles that have shallow cuts, because the intersection of two distinct faces has few pieces and thus a basic 4-move commutator affects a small number of pieces. From there, we can build more complicated commutators to get 3-cycles and whatnot. If you can set up any three pieces (3-transitivity), then the moment you find a 3-cycle commutator (or even a 3-cycle alg that isn't a commutator), you can automatically get every even permutation.

Also, as alluded to in the Puzzling SE question, odd permutations can never be handled with commutators, which (in a permutation group) are always even permutations. More generally, we have the concept of the

commutator subgroup, which is the subgroup generated by every commutator. We can consider the quotient group given by the puzzle's group modulo the commutator subgroup. This quotient group is always abelian (since commutators measure the extent to which commutativity fails, and you've modded out by all of them), so you could "solve the puzzle" in this abelian quotient group first, then use commutators for what's left. E.g. for a 4×4×4 supercube, the quotient group is isomorphic to \( (\mathbb Z/2)\times(\mathbb Z/2) \), corresponding to the parity of the wings and the parity of the corners; if you fix those first (with quarter turns of an inner slice or outer slice, respectively), then the resulting puzzle state can be solved with commutators only.

So I guess the hidden question here might be: what's an example of a twisty puzzle with a group structure such that the abelianisation has some interesting structure to it, rather than just being parity or something similar? (Let's count the pyraminx "centres" as falling under the something-similar category, even though of course that's not really parity.)