The "generalized dihedral group" for an abelian group A is the semidirect product of A and a cyclic group of order two acting via the inverse map on A. A thus has index two in the whole group and all elements outside A have order two. Thus, at least half the elements of any generalized dihedral group have order two.
My question is the converse: if half or more the elements of a finite group G have order two, is it necessary that G is either an elementary abelian 2-group or a generalized dihedral group? [Note: Actually nontrivial elementary abelian 2-groups are also generalized dihedral, they're an extreme case. Also, note that the direct product of a generalized dihedral group with an elementary abelian 2-group is still generalized dihedral, because the elementary abelian 2-group can be pulled inside the abelian part.]
I have a proof when the order of G is twice an odd number m. In that case, there is a short an elegant elementary proof -- we consider the set of elements that can be written as a product of two elements of order two and show that this is a subgroup of order m, then show that any element of order two acts by the inverse map on it, and hence the subgroup is abelian. It can also be thought of as a toy example of Frobenius' theorem on Frobenius subgroups and complements (though we don't need Frobenius' theorem).
However, I am having some difficulty generalizing this, mainly because there are elements of order two that are inside the abelian group.
Although I have a number of possible proof pathways, I'll refrain from mentioning them for now because the actual proof is likely to be much simpler and I don't want to bias others trying the problem.