This answer includes an application of the 4th isomorphism theorem for rings, and one (well, kinda two) of the 2nd isomorphism theorem for groups. I also link to a really cool application of the 4th isomorphism theorem in groups (but is kinda the same as the application to rings).
Rings, 4th isomorphism theorem
An neat application of the 4th isomorphism theorem for rings is the following:
Theorem: Maximal ideals are prime.
Proof: Let $R$ be a ring and let $I$ be a maximal ideal of $R$. Then consider the quotient $R/I$, and apply the correspondence theorem.
Indeed, if $R$ is commutative with identity then $R/I$ is a field. See here for a proof.
Groups, 4th isomorphism theorem
There is a rather cool trick in the theory of infinite groups, which was used by Higman to construct an infinite simple group. The idea is to appeal to Zorn's lemma and obtain a maximal normal subgroup, and then quotient this out to get a simple group. See here for a neat application of it. This is very similar to the above application to rings. (You quotient out a maximal subgroup/ideal to get a simple group/prime ring.)
Groups, 2nd isomorphism theorem
A nice application of the 2nd isomorphism theorem for groups is the following. It deals with soluble groups, which are an important class of groups (and are often taught in a second course on groups). A group $G$ is soluble (or solvable) if it possesses an abelian series, that is, a series $$1=G_0\lhd G_1\lhd\cdots\lhd G_n=G$$ where each factor $G_{i+1}/G_i$ is abelian.
Theorem: Subgroups and homomorphic images of soluble groups are soluble.
Proof: Suppose $G$ is soluble with abelian series $1=G_0\lhd G_1\lhd\cdots\lhd G_n=G$.
Subgroups: If $H$ is a subgroup of $G$ then we can apply the 2nd isomorphism theorem to get the following.
$$\frac{H\cap G_{i+1}}{H\cap G_i}\cong\frac{(H\cap G_{i+1})G_i}{G_i}\leq\frac{G_{i+1}}{G_i}$$
Hence, the groups $\frac{H\cap G_{i+1}}{H\cap G_i}$ are abelian so the set $\{H\cap G_i; i=0, 1, \ldots, n\}$ forms an abelian series for the group $H$, as required.
Homomorphic images: If $N$ is a normal subgroup of $G$ then the can apply the 2nd isomorphism theorem to get the following.
$$\frac{G_{i+1}N}{G_iN}\cong\frac{G_{i+1}}{G_{i+1}\cap (G_iN)}$$
The subgroup $G_i$ is a subgroup of $G_{i+1}$ and of $G_iN$, and so $G_i\lhd G_{i+1}\cap (G_iN)$. Therefore, $\frac{G_{i+1}}{G_{i+1}\cap (G_iN)}$ is a homomorphic image of $\frac{G_{i+1}}{G_i}$, and hence is abelian. By the 3rd isomorphism theorem we have the following.
$$\frac{G_{i+1}N/N}{G_iN/N}\cong \frac{G_{i+1}N}{G_iN}$$
Hence, the set $\{G_iN/N; i=0, 1, \ldots, n\}$ forms an abelian series for $G/N$, as required.
Below, "ring" means "unital ring."
In the comments above, runway44 has pointed out that there is in fact a huge gap between a full ring structure on $G$ and a choice of multiplicative identity. However, I'm going to focus on the question you've asked specifically:
is every element of an arbitrary abelian group $G$ a candidate for $1_G$ when defining a ring structure? If so, what are the consequences of different choices, and if not what are the necessary group-theoretic requirements?
Given an abelian group $G$, let $I(G)$ be the set of elements $a$ of $G$ such that there is a ring $R$ with additive group $G$ and multiplicative identity $a$. At a most basic level, we just want to compute $I(G)$ given $G$.
Let's start with a triviality. If $G$ is nontrivial then $I(G)$ is not all of $G$ since the additive identity is also a multiplicative annihilator in any ring. Additionally, there are some "un-ringable" abelian groups - that is, nontrivial groups satisfying $I(G)=\emptyset$. The simplest example of such in my opinion is $\mathbb{Q}/\mathbb{Z}$.
Now we come to a slightly less trivial observation: we can in fact have $\emptyset\subsetneq I(G)\subsetneq G\setminus\{e_G\}$. Given a ring $R$ and an element $r\in R$, let $ord_R(r)$ be the order of the cyclic subgroup of the additive group of $R$ generated by $r$. For example, if $R$ is a field and $r$ is the multiplicative unit, then $ord_R(r)$ is just the characteristic. Distributivity implies that $ord_R(r)\vert ord_R(1_R)$ for every $r\in R$. Turning this around, we get a constraint on the function $I$ defined above: for example, every element of $I(\mathbb{Z}/6\mathbb{Z})$ must have order $6$ (and since all elements of order $6$ in $\mathbb{Z}/6\mathbb{Z}$ are automorphic, this gives a complete description of $I(\mathbb{Z}/6\mathbb{Z})$).
Best Answer
The answer is "yes".
Every ring is a homomorphic image of the free ring $\mathbb{Z}\langle X\rangle$ generated by some sufficiently large set $X$. So it suffices to prove that $\mathbb{Z}\langle X\rangle$ is the endomorphism ring of some abelian group.
There may be a much more elementary way to show this, but it follows from the main theorem of
Dugas, Manfred; Göbel, Rüdiger, Every cotorsion-free algebra is an endomorphism algebra, Math. Z. 181, 451-470 (1982). ZBL0501.16031,
which (specialized from the more general context of modules for a Dedekind domain $R$ to the case $R=\mathbb{Z}$) states that every ring whose additive group is a cotorsion-free abelian group is the endomorphism ring of some abelian group.
Since the additive group of $\mathbb{Z}\langle X\rangle$ is a free abelian group, and therefore cotorsion-free, it follows that $\mathbb{Z}\langle X\rangle$ is the endomorphism ring of some abelian group.