Is this an equivalent formulation of a ring? Let $G$ be an abelian group such that there exist two families $\{\psi_g, \phi_g\}_{g \in G}$, not neccessarily disjoint, of homomorphisms $\psi_g, \phi_g : G \to \langle g \rangle$, each mapping onto the corresponding cyclic subgroup; such that $\phi_g(h) = \psi_h(g)$ for all $g,h \in G$.  Then we have a ring $G$ with $\cdot$ defined by $g \cdot h = \psi_g(h)$.
This seems right since distributivity is replaced by homomorphism since distributivity looks like a homomorphism.

Associativity is achieved when we also impose the requirement that $\psi_{g \cdot h} = \psi_g \circ \psi_h$.  In other words the induced multiplication is compatible with $G$ endomorphism composition.

So take the additive integers.  Clearly $\psi_a(x) = ax$ satisfies these conditions, and the ring you arrive at is $\Bbb{Z}$.
 A: There is some part of your proposal that works, and I try to explain here. 
For any ring $R$, denote by $\operatorname{End}_{\mathbb Z}(R)$ the ring of endomorphisms of $R$ as $\mathbb{Z}$-module = abelian group. It is a ring with the addition of morphisms and the composition. 
Now, you can consider the map $$\rho: R \to \operatorname{End}_{\mathbb Z}(R)$$ given by sending $a$ to the endomorphism $\rho(a)(b):=a\cdot b$ (note that $\rho(a)$ is essentially what you called $\psi_a$). This gives you a morphism of rings, since $$\rho(a\cdot b)(c)=(a\cdot b)\cdot c=a\cdot (b\cdot c)=\rho(a)(\rho(b)(c))=(\rho(a)\circ \rho(b))(c)$$
(so, by the associative property). 
Conversely, given any abelian group $G$, and a morphism of abelian groups 
$\psi: G\to \operatorname{End}_{\mathbb Z}(G)$, you can define $a\odot_\psi b:=\psi(a)(b)$. In order to get a ring with this operation, the good thing is that the distributive properties come from the fact that $\psi$ is a morphism of abelian groups (on one side), and that we are considering elements on $\operatorname{End}_{\mathbb Z}(G)$ (for the other side): 
$$(a+b)\odot_\psi c=\psi(a+b)(c)=\psi(a)(c)+\psi(b)(c)=a\odot_\psi c+b\odot_\psi c$$
$$a\odot_\psi (b+c)=\psi(a)(b+c)=\psi(a)(b)+\psi(a)(c)=a\odot_\psi b+a\odot_\psi c$$
But you need to impose, first, that there is an element $1\in G$ that goes to the identity (so the unit) and verifies $\psi(a)(1)=a$ for all $a\in G$, and the associative property:
$$a\odot_\psi (b\odot_\psi c)=\psi(a)(\psi(b)(c))=(\psi(a)\circ \psi(b))(c)$$
should be equal to 
$$(a\odot_\psi b)\odot_\psi c=\psi(\psi(a)(b))(c),$$
so the following property should hold 
$$\psi(a)\circ \psi(b)=\psi(\psi(a)(b))$$
for all $a, b\in G$. 
A: It is not true in general that, if $R$ is a ring, and $a\in R$, for any $b\in R$, $b\cdot a\in \langle a \rangle$. Thus "your" definition of ring only works if $R$ is cyclic (as abelian group), and thus you get $R\cong \mathbb Z $ or  $R\cong \mathbb Z / n\mathbb Z $ as rings. 
