Maybe the socle itself isn't that important, but its constituent parts, the minimal normal subgroups, are very important, particularly in soluble groups.
Here is one vitally important example. Let $G$ be a primitive permutation group (i.e., transitive and the point stabilizer is a maximal subgroup). If $G$ is soluble, then the degree of $G$ is a power $p^n$ of a prime $p$. This degree is the order of some minimal normal subgroup, which since $G$ is soluble, must be an elementary abelian $p$-group (direct product of cyclic groups of order $p$). This subgroup is regular on the set.
The socle appears in primitive permutation groups, because of the O'Nan--Scott theorem, which sets out to list all primitive permutation groups in some sense (it originally classified the maximal subgroups of symmetric groups, but has been adapted). It's important to know whether there is more than one minimal normal subgroup (i.e., whether the socle is minimal normal or a product of two).
Another use of minimal normal subgroups is in reduction theorems for finite groups. Let $G$ be an arbitrary finite group, and suppose you want to prove some conjecture about it. You have the classification of finite simple groups, and you want to apply that. This means you need to reduce the problem to a question about simple groups.
For some problems it is often easy to eliminate the case where there is an abelian normal subgroup. Thus a minimal normal subgroup is a direct product of isomorphic simple groups. The socle of this group, often called the layer or the Bender subgroup in this situation, contains its own centralizer. (Note that this requires there to be no abelian normal subgroups!) If $N$ is a normal subgroup of $G$ such that $C_G(N)\leq N$ and $N$ is a product of non-abelian simple groups, then $C_G(N)=1$ and so $G$ is a subgroup of $\mathrm{Aut}(N)$.
Another example is with classifying maximal subgroups of finite groups. A theorem of Aschbacher and Scott states that one can understand the maximal subgroups of all finite groups if and only if one can understand the maximal subgroups of groups $M\rtimes G$, where $G$ is a simple group (technically, almost simple, but simple is interesting enough), and $M$ is an elementary abelian $p$-group for some prime $p$ on which $G$ acts. Here the socle is $M$, and this reduction theorem was obtained by looking precisely at this sort of minimal normal structure.
As noted, and as you know, every subgroup of an abelian group is normal; so the socle of an abelian group $G$ is generated by the minimal nontrivial subgroups.
If $H$ is a minimal nontrivial subgroup of $G$, and $h\in H$, $h\neq 1$, then $\langle h\rangle = H$; thus, $H$ is cyclic, and has no nontrivial subgroups. Therefore, $H$ is cyclic of prime order. Conversely, if $H$ is a subgroup of prime order, then it is necessarily minimal nontrivial, since it has no proper nontrivial subgroups.
So the socle must be generated by the elements of prime order. In particular, $(\mathbb{R},+)$, being torsionfree, has trivial socle, as does every torsionfree abelian group.
Thus, if $G$ is abelian, then $\mathrm{socle}(G)=\mathrm{socle}(G_{\rm tor})$, the torsion subgroup of $G$. And
$$\mathrm{socle}(G) = \langle x\mid x\text{ has prime order}\rangle.$$
You can get a bit more information (depending on how much Robinson has proven). Since the torsion subgroup is the direct sum of its $p$-parts, you can just look at the separate $p$-parts, and then just look at what some authors denote $G[p]$, the set of elements of order dividing $p$. So
$$\mathrm{socle}(G) = \bigoplus_{p\text{ prime}}G[p].$$
(Your final example comes from the fact that $\mathbb{Z}_6\cong\mathbb{Z}_2\times\mathbb{Z}_3$, so you "really" have $\mathbb{Z}_2\times\mathbb{Z}_3^2$...)
Best Answer
I think the equality is true. Here is a sketch proof.
Let $N$ be a minimal normal subgroup of $G = A_1 \times \cdots \times A_n$. If $N \le A_i$ for some $i$ then $N \le {\rm Soc}(A_i)$, so suppose not, and suppose that $N$ projects nontrivially onto $A_1$ and $A_2$, say.
Then $N$ cannot contain any elements $(h,1,\ldots)$ with $h\ne 1$ because the set of such elements would form a smaller normal subgroup than $N$.
But this implies that the projection $N_1$ of $N$ onto $A_1$ must lie in $Z(A_1)$, since otherwise $N$ would contain elements $([g,h],1,\ldots)$ with $h \in N_1$, $g \in A_1$ and $[g,h] \ne 1$.
So $N \le Z(G)$, and $N$ must have prime order, and be a diagonal subgroup of minimal normal subgroups of some of the $A_i$. So $N \le {\rm Soc}(A_1) \times \cdots \times {\rm Soc}(A_n)$