Answer:
I cheated and asked Richard Lyons this question (or at least, the reformulation of the problem, conjecturing that (G,c) is nonsplitting for an involution c with <c> generating G if and only if there exists an odd A such that G/A = Z/2). His response:
Good question! This is a famous (in my circles) theorem - the Glauberman Z^*-Theorem. (Z^*(G) is the preimage of the exponent 2 subgroup of the center of G/O(G), and O(G)=largest normal subgroup of G of odd order.)
Z^*-Theorem: If c is an involution of G then c\in Z^*(G) iff [c,g] has odd order for all g\in G iff for any Sylow 2-subgroup S of G containing c, c is the unique G-conjugate of itself in S.
The last property is absolutely fundamental for CFSG. The proof uses modular character theory for p=2. Attempts to do it with simpler tools have failed.
George Glauberman, Central Elements in Core-free Groups, Journal of Algebra 4, 1966, 403-420.
Older Remarks:
Comment 1: Suppose that P = Z/2+Z/2 is a 2-Sylow. If x lies in P, then P clearly centralizes x, and thus the order of <x> divides #G/P, and is thus odd. By a theorem of Frobenius, G has an odd number of elements of order 2, and thus we see it has an odd number of conjugacy classes of elements of order 2. Yet, by the Sylow theorems, every element of order 2 is conjugate to an element of P. If c lies in P, then by nonsplitting, it is unique in its G-conjugacy class in P. Thus there must be exactly three conjugacy classes of elements of order 2, and thus no element of P is G-conjugate. By a correct application of Frobenius' normal complement theorem, we deduce that G admits a normal subgroup A such that G/A \sim P. Yet <c> generates G, and thus the image of <c> generates G/A. Yet G/A is abelian and non-cyclic, a contradiction.
Comment 2: Suppose that A is a group of order coprime to p such that p | #Aut(A).
Let G be the semidirect product which sits inside the sequence:
1 ---> A ---> G --(phi)--> Z/pZ --> 0;
Let c be (any) element of order p which maps to 1 in Z/pZ. If c is conjugate to
c^j, then phi(c) = phi(c^j). Hence c is not conjugate to any power of itself.
Let H be a subgroup of G containing c (or a conjugate of c, the same argument applies). The element c generates a p-sylow P
of H (and of G). It suffices to show that if gcg^-1 lies in H, then it is conjugate to c inside H. Note that gPg^-1 is a p-Sylow of H.
Since all p-Sylows of H are conjugate, there exists an h such that gPg^-1 = hPh^-1, and thus h c^j h^-1 = gcg^-1. Yet we have seen that c^j is not conjugate to c inside G unless j = 1. Thus gcg^-1 = hch^-1 is conjugate to c inside H.
I just noticed that you wanted <c> to generate G. It's not immediately clear (to me) what condition on A one needs to impose to ensure this. Something like the automorphism has to be "sufficiently mixing". At the very worst, I guess, the group G' generated by <c> still has the property, by the same argument.
This works more generally if p || G and no element of order p is conjugate to a power of itself. (I think you know this already if p = 2.)
The case where the p-Sylow is not cyclic is probably trickier.
Examples: A = (Z/2Z)+(Z/2Z), p = 3. (This is A_4).
A = Quaternion Group, p = 3. (This is GL_2(F_3) = ~A_4, ~ = central extension).
A = M^37, M = monster group, p = 37.
Best Answer
My standard rant about "what can we say about $G$": what we can say about $G$ is that the two partitions are the same. If the questioner doesn't find that a helpful answer then they might want to consider the possibility that they asked the wrong question ;-)
But as to the actual question: "is $G$ forced to be abelian?", the answer is no, and I discovered this by simply looping through magma's database of finite groups. Assuming I didn't make a computational slip, the smallest counterexample has order 64, is the 73rd group of order 64 in magma's database, which has 8 representations of degree 1, 14 representations of degree 2, 8 elements in the centre and 14 more conj classes each of order 4.
Letting the loop go further, I see counterexamples of size 64, 128, 192 (I guess these are just the counterexamples of size 64 multiplied by Z/3Z) and then ones of order 243 (a power of 3). So I guess all examples I know are nilpotent. Are they all nilpotent? That's a question I don't know the answer to.