Commutative ring and its group-algebra, and abelian-group-algebra as a commutative ring. In course of discussing the algebraic structures, one of my seniors is led quite naturally to considering the $\color{red} {geometric}$ version of following:

Question: Since we could consider the spectrum of an abelian-group-algebra $\color{blue}{as\ a\ commutative\ ring}$, it is natural to ask: what kind of informations of this abelian group could be obtained from the considerations of this spectrum?
  On the other hand, we can view the commutative ring $\color{blue} {as\ an\ abelian\ group}$, and thus form its group-algebra. We then ask a further question: what relations are there between the spectrum of the ring, and the spectrum of this group-algebra?
Notice: As I have yet heard of nothing about the subject, any source or reference in this direction is the most appreciated. Thanks in advance.  

Also one of my seniors is greatly intrigued in the $\color{green}{geometric}$ point of view of this question, thus it would be quite wonderful if any insight or crucial observations are provided. Thanks again.
 A: The multiplicative structure of rings is crucial for doing algebraic geometry. In fact, most of it can be done for arbitrary commutative monoids (see the work by Connes, Deitmar, etc.), and more generally, for commutative algebraic monads (see Durov's thesis).
I don't think that the additive structure carries enough information about a ring in order to say something interesting. In particular, the functor $\mathsf{CRing} \to \mathsf{Ring}, (A,+,*) \mapsto \mathbb{Z}[A,+]$ loses lots of information. And in general there are no ring homomorphisms $\mathbb{Z}[A] \to A$ or $A \to \mathbb{Z}[A]$ at all.
EDIT: If $A$ is just an abelian group (not assumed to be the underlying additive group of a ring), then $\mathbb{Z}[A]$ is a quite well-understood commutative ring (in particular all the open problems concerning zero divisors etc. in group rings are solved in this case). It is the directed colimit of the group rings $\mathbb{Z}[A']$, where $A'$ runs through the finitely generated subgroups of $A$. Now, if $A$ is finitely generated, then there is an isomorphism $A \cong \mathbb{Z}^r \oplus \mathbb{Z}/n_1 \oplus \dotsc \oplus \mathbb{Z}/n_s$ for integers $r,s$ and $n_1 | \dotsc | n_s>0$. It follows $\mathbb{Z}[A] \cong \mathbb{Z}[x_1^{\pm 1},\dotsc,x_r^{\pm 1}][y_1,\dotsc,y_s]/(y_i^{n_i}-1)_i$. In particular, we see that if $A$ is torsion-free, then $\mathbb{Z}[A]$ is a regular integral domain.
