I already proved $R$ is a subring, but it's the local bit that's giving me trouble. I believe I have it, but I would appreciate a second (or eighth) eye.
Assume $R$ is not local. So $R$ has more than one maximal ideal. Let two of these maximal ideals be $M$ and $N$, $M \neq N$.
Let $(a,b) = \frac{a}{b} \in M$. $p$ does not divide $b$.
$(a,b)$ is contained in an ideal, so it cannot be a unit. So $p$ must divide $a$ (if $p$ doesn't divide $a$, $(b,a) \in R$, and $\frac{a}{b} \frac{b}{a} = 1 \notin M$).
So $(a,b) \in \frac{(p)}{\mathbb Z - (p)}$.
$(p)$ is a maximal ideal of $\mathbb Z$, so $\frac{(p)}{\mathbb Z - (p)}$ is a maximal ideal of $\frac{\mathbb Z}{\mathbb Z - (p)} = R$ since $p$ is prime.
But every maximal ideal of $\mathbb Z$ is of the form $(q)$, $q$ is prime. So for $N \neq M$, $N = \frac{(q)}{\mathbb Z - (p)}$, $q \neq p$.
But $p$ does not divide $q$, so for any $(nq,b) \in R$, $(b,nq) \in R$, where $nq$ is a multiple of $q$ and $b$ is an integer. So $N = \frac{(q)}{\mathbb Z - (p)}$ are all units, and therefore cannot make up an ideal, maximal or not.
Therefore $M$ is the only maximal ideal of $R$, and $R$ is local.