I think I have this question figured out almost completely, but I'm a little worried about using a certain notation.
Suppose $\mathbb{Z} \stackrel{\phi}{\longrightarrow} R$ is a ring homomorphism. Observe that $\phi(1) = \phi(1\cdot1) = \phi(1)\phi(1) = \phi(1)^2$. Thus, $\phi(1)-\phi(1)^2 = 0$ and moreover $\phi(1)(\phi(1) - 1) = 0$. Since R has no zero divisors, it follows that $\phi(1) = 0$ or $\phi(1) = 1$. Hence, $\phi$ is either trivial, or $\phi(1) = 1$. In the latter case, $\phi$ is entirely determined in the following way: for any $r \in \mathbb{Z}$, \begin{equation} \phi(r) = \phi(\stackrel{r \, \,times}{1 \, + \, ... \, + \, 1}) = \, \stackrel{r \, \,times}{\phi(1) \, + \, ... \, + \, \phi(1)} = r\cdot\phi(1) \end{equation}
My trepidation arises with the last equality. It doesn't seem to me that multiplication of an integer r by $\phi(1)$ is necessarily well defined in an arbitrary integral domain R. That being said, I don't see any other way to denote iterated addition. Is this a well-defined notation, and if not, how should I represent iterated addition in R?
If $R$ is a ring with unity everything works fine. You have an action of $\mathbb{Z}$ on $R$ namely if $m\in \mathbb{Z}$ we define $mr$ as you said above.
This is because every ring has an abelian group structure with the sum.
Returning to your question, if you avoid a morphism that sends everything to zero the rest of homomorphisms are determined by the image of 1.