Amongst the finitely generated abelian groups, those with commutative endomorphism ring are exactly the cyclic groups.
Torsion abelian groups with commutative endomorphism rings are exactly the locally cyclic groups, that is, the subgroups of Q/Z. They were classified in:
- Szele, T.; Szendrei, J.
"On abelian groups with commutative endomorphism ring."
Acta Math. Acad. Sci. Hungar. 2, (1951). 309–324
MR51835
DOI:10.1007/BF02020735
This papers also gives more complicated examples of mixed groups with commutative endomorphism ring.
The mixed case was completed in:
- Schultz, Ph.
"On a paper of Szele and Szendrei on groups with commutative endomorphism rings."
Acta Math. Acad. Sci. Hungar. 24 (1973), 59–63.
MR316598
DOI:10.1007/BF01894610
This paper indicates the difficulty of any classification of torsion-free abelian groups with commutative endomorphism rings, as Corner has shown that very large torsion-free abelian groups can have commutative endomorphism rings (while the classifications up to now have basically been "only very small ones").
If $G$ is a finite abelian group, then $\mathbb C[G] = \lbrace f \colon \hat G \to \mathbb C \rbrace$, where $\hat G$ is the Pontrjagin dual of $G$. The isomorphism $g \mapsto g^{-1}$ translates into the same map on the Pontrjagin dual (basically multiplication by $-1$ on $\hat G$), but now it is a bit easier to analyze. Note also, that there is a non-canonical isomorphism $G \cong \hat G$.
Hence, in order to find two non-isomorphic abelian groups which have strongly isomorphic complex group rings, we just have to analyze (in addition to the cardinality of the group) the orbit structure of multiplication by $-1$ on the group. Indeed, we are now just talking about an algebra of function on a set with some $\mathbb Z/2 \mathbb Z$-action.
Example: In $\mathbb Z/8\mathbb Z \times \mathbb Z/2\mathbb Z$, there are precisely $4$ elements, which are fixed under multiplication by $-1$, namely $(0,0),(4,0),(0,1)$ and $(4,1)$. The same is true for $\mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z$. Here, one has $(0,0),(2,0),(0,2)$ and $(2,2)$. Hence, there exists an isomorphism between $\mathbb C[\mathbb Z/8\mathbb Z \times \mathbb Z/2\mathbb Z]$ and $\mathbb C[\mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z]$, which respect the isomorphism which is induced by $g \mapsto g^{-1}$.
I do not know about an example with coefficients in $\mathbb Z$.
Best Answer
The rings, you call ``complete'' are known as $E$-rings (as Ulrich Pennig mentioned in the comments).
Some comments on your questions