Finding a compatible multiplication for a given group
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.
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. \\\