I didn't quite finish a characterization, but certainly such a group must be a semidirect product of an abelian group with a product of $\mathbb{Z}/2\mathbb{Z}$'s.
If $G$ has $n$ 1-dimensional quaternionic representations whose sum is faithful, then $G$ must embed into the product of $n$ number of copies of the unit quaterions (that is, $SU(2)$). Consider the image of $G$ in any one of these representations; since it is a finite subgroup of $SU(2)$, it is either cyclic, dihedral, or a central extension of $A_3,A_4$ or $A_5$ (see Wikipedia). However, if it's one of the latter 3, this group has a different irrep over the quaternions that isn't 1-d, so it must be cyclic or dihedral.
Thus, $G$ is a subgroup of a product of cyclic and dihedral groups. The intersection of $G$ with the product of the normal cyclic subgroups inside the dihedrals (and all of the cyclic group factors) gives a normal subgroup of $G$, and the quotient of $G$ by this subgroup is a product of $\mathbb{Z}/2\mathbb{Z}$'s. Also, this quotient is split by just taking any preimage of the each of the generators of the product of $\mathbb{Z}/2\mathbb{Z}$'s. Thus, we have a semi-direct product.
I suspect you can use some character calculations to see that actually this is a product of cyclics and dihedrals, but it's not working out at the moment, and I wouldn't be totally shocked if there are some funny exceptions.