Is there a theory or characterization for those finite $p$groups that can be considered as the additive group of a finite local commutative ring with identity?
Any nonzero finite abelian $ p $  group is the additive group of a commutative local ring.
Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ \mathbb{Z}/p^k \mathbb{Z} \times M $ where $ p^k = \exp (G) $. Then $ M $ naturally has the structure of a $ \mathbb{Z}/p^k \mathbb{Z} $  module.
Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.
Now let's prove it's local with maximal ideal $ \{ (x,m) \mid p $ divides $x \} $.
Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, y^2m) = (xy, xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.
This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $  module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.

2


$\begingroup$ What is the $M$ in $G = \mathbb Z/p^k\mathbb Z \times M$? $\endgroup$ – LSpice Apr 24 at 18:40

3$\begingroup$ @LSpice : you can see it as $G/(\mathbb{Z}/p^k\mathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups $\endgroup$ – Max Apr 24 at 19:06

$\begingroup$ @Max, thanks. I missed your clarification that $p^k = \exp(G)$, so that $M$ is just "what's left". $\endgroup$ – LSpice Apr 26 at 1:25