Finding a compatible multiplication for a given group If you are given an abelian group $\ (G, +)$, is there some algorithm to find all possible semigroups $\ (G, ×)$, such that $\ (G, +, ×)$ is a ring?
If not, can you at least decide, if the ring must have zero divisors?
 A: First, let's deal with the case when $(G,+)$ is finitely generated.  By the fundamental theorem of finitely generated abelian groups, let's go ahead and assume that $G$ is given to us in the form $\bigoplus_{i=1}^{m} \mathbb{Z}/n_i\mathbb{Z}$ where $n_1|n_2|\cdots |n_m$ are non-negative integers.
  [Some of them could be zero.]  Let $g_i$ be a generator of $\mathbb{Z}/n_i\mathbb{Z}$.
There are at most countably many choices for $g_i g_j\in G$, which we can easily enumerate. Thus, an algorithm can run through each possibility.  So we are reduced to the following question: Given a table of $m\times m$ elements of $G$ (where we think of the $(i,j)$ value as $g_i g_j)$ can we decide whether this gives a ring structure to $G$ (by extending in the obvious way, using addition and distributivity, to a multiplication on $G$)?  I haven't worked out all the details, but this seems plausible to me.  Checking whether or not this multiplication is well-defined, associative, has a unit, and is distributive should all turn into solvable linear algebra problems.  I'll leave it to you to double-check that this works.  [Sketch: Define $$(a_1 g_1 + a_2 g_2 + \cdots + a_m g_m)(b_1 g_1 + b_2 g_2 + \cdots + b_m g_m) := \sum_{i,j} a_i b_j \varphi(i,j),$$ where $\varphi(i,j)$ is the $(i,j)$ value in your table.  To check well-definedness, replacing $a_i$ by $a_i+k n_i$, we need to check that the new resulting output is congruent (modulo $n_j$ in the $j$th coordinate, for each $j$) to the original output (repeating the process with the $b$'s as well).  etc...]
Second, when $(G,+)$ is not finitely generated, there are all sorts of problems.  One is that I don't know how you'd tell the computer which group you are talking about.  Another is that a countable group can have uncountably many compatible ring structures [just take $G$ to be a countable dimensional $\mathbb{F}_2$-vector space, for instance] and so I don't know how an algorithm would output uncountably many ring structures for you.
A: If not, can you at least decide, if the ring must have zero divisors?
It would be interesting to know if there is a good answer to this question. Here I will just make three remarks. I will call an abelian group $G$ good if it can be equipped with a bi-additive multiplication which makes it a ring with no nontrivial zero divisors.


*

*By this answer, there is an abelian group $G$ that is elementarily equivalent to the group of integers $\mathbb Z$ such that the only bi-additive operation on $G$ is the zero function. This means that $\mathbb Z$ is good, $G$ is not, yet $\mathbb Z$ and $G$ satisfy the same first-order sentences. 

*If $G$ is good and has a nonidentity element $g$ of finite additive order, then $G$ must be an elementary abelian $p$-group for some prime $p$. Conversely, any elementary abelian $p$-group is good. [For the first assertion, if $m\in\mathbb Z$, $g, h\in G-\{0\}$, and $mg=0$, then $0=(mg)h=g(mh)$, so by the goodness of $G$ we get $mh=0$. This shows that once one nonidentity element of $G$ has finite order, then all nonidentity elements have the same finite order, which must be prime. For the second assertion, any elementary abelian $p$-group supports a field structure, hence is good.]

*This remark addresses the first interesting cases where the structure of $G$ is not decided by Remark 2, which are the cases where $G$ is torsion-free of rank one.  
Claim. If $G$ is a torsion-free abelian group of rank one, then $G$ is good  iff $G$ is isomorphic to the additive group of a unital subring of $\mathbb Q$.
Besides being a torsion-free abelian group of rank one, the additive group of any unital subring of $\mathbb Q$ has the following additional property: Call an element $g\in G$ $m$-divisible for a given positive integer $m$ if the equation $mx=g$ is solvable in $G$. Call $G$ $m$-divisible if every $g\in G$ is $m$-divisible. The additional property is: there is an element $u\in G$ such that for every prime $p$ we have ($u$ is $p$-divisible implies $G$ is $p$-divisible).
Pf. 
The if part of the claim is clear: the additive subgroup of a unital subring of $\mathbb Q$ is a good, torsion-free, abelian group of rank one.
It is easy to see why the additive groups of unital subrings of $\mathbb Q$ satisfy the additional property. If $G$ is the additive group of a unital subring of $\mathbb Q$, choose $u=1\in G$. Then if $px=1$ is solved by $x=x_p$, for any $g\in G$ we get that $py=g$ is solved by $y=x_p\cdot g$. 
Here is a sketch for why the additional property characterizes the subgroups of unital subrings of $\mathbb Q$ among all torsion-free abelian groups of rank one. If $G$ is torsion-free of rank one, then there is a group embedding $\varphi:G\to\mathbb Q$. If $u\in G$ is an element such that for every prime $p$ we have ($u$ is $p$-divisible implies $G$ is $p$-divisible), then we can scale the embedding to $\frac{1}{u}\varphi:G\to Q$ so that the image of $u$ is $1$. Now the property that for every prime $p$ we have ($u$ is $p$-divisible implies $G$ is $p$-divisible) translates into the property that the set $\frac{1}{u}\varphi(G)$ consists of all rational numbers whose denominators are divisible only by those primes for which $u$ is $p$-divisible. This is a unital subring of $\mathbb Q$.
What remains to explain is why if $G$ is torsion-free of rank one, and $G$ contains no element $u$ such that ($u$ is $p$-divisible implies $G$ is $p$-divisible), then $G$ is not good.
Argument:
This is a proof by contradiction, so assume that 
$G$ is torsion-free of rank one, and $G$ contains no element $u$ such that ($u$ is $p$-divisible implies $G$ is $p$-divisible), YET $G$ is nevertheless good.
Applying an embedding if necessary, assume also that $G$ is an additive subgroup of $\mathbb Q$ that contains $1$. Suppose that $G$ is NOT $p$-divisible for exactly the primes $p_1<p_2<\cdots$. Using the facts that $1\in G$ and $G$ is not $p_i$-divisible, conclude that there is a nonnegative integer $e_i$ such that $\frac{1}{p_i^{e_i}}\in G$, but no reduced fraction $\frac{r}{s}\in G$ has denominator divisible by $p_i^{e_i+1}$.
I claim that the $e_i$'s will be positive infinitely often. For if there is some $k$ such that $e_i=0$ when $i>k$, then 
$u=\frac{1}{p_1^{e_1}\cdots p_k^{e_k}}\in G$ is an element that is not $p$-divisible for the same primes as $G$ ($p_1<p_2<\dots$), and we have assumed that there is no such element.
Choose any $i$ where $e_i>0$. 
Let $\ast$ denote a bi-additive multiplication on $G$ that witnesses that $G$ is good. Write $1\ast 1$ as $\frac{r}{s}$ in reduced form, and write 
$\frac{1}{p_i^{e_i}}\ast \frac{1}{p_i^{e_i}}$ as $\frac{r_i}{s_i}$ in reduced form. Since $1 = p_i^{e_i}\cdot \frac{1}{p_i^{e_i}}$ and $\ast$ is bi-additive, 
$$\frac{r}{s}=1\ast 1 = p_i^{2e_i}\cdot \left(\frac{1}{p^{e_i}}\ast \frac{1}{p^{e_i}}\right) = p^{2e_i}\frac{r_i}{s_i}.$$
Since no reduced fraction in $G$ has denominator divisible by $p_i^{e_i+1}$, it follows that the numerator $r$ of $\frac{r}{s} = \frac{p^{2e_i}r_i}{s_i}$ is divisible by $p_i^{e_i}$. As $i$ ranges we see that the fixed integer $r$ is divisible by infinitely many primes, a contradiction.  \\\
