the localization of a ring is an integral domain iff the annihilators of zero divisors are comaximal ideals

I suppose your approach is correct, but you need to be a little bit more delicate.

Let us begin with a observation concerning localization in general:

Lemma

If $\mathfrak{p}$ is a prime ideal, then for any $x \in R$ such that $x/1 \neq 0 \in R_\mathfrak{p}$ then $\mathrm{Ann}(x) \subset \mathfrak{p}$.

Proof

Indeed, suppose that $t \in \mathrm{Ann}(x)$, or in other words $tx = 0$ and $t \not \in \mathfrak{p}$. Then, $1/t$ is a valid element of $R_\mathfrak{p}$ and we can compute: $$ 0 = tx/1 \cdot 1/t = tx/t = x/1 \neq 0 $$ which forms the desired contradiction. $\square$

Now, suppose as you did, that $x/s \cdot y/t = 0 \in R_\mathfrak{p}$. Then you rightly conclude that $xyz = 0$ for some $z \not \in \mathfrak{p}$. It will suffice to prove that $x/1 = 0$ or $y/1 = 0$. Suppose otherwise, namely that $x/1 \neq 0$ and $y/1 \neq 0$. Then, we argue that also $xz/1 \neq 0$, since $z/1$ is invertible. By the lemma, we know that $$\mathrm{Ann}(xz), \mathrm{Ann}(y) \subset \mathfrak{p}$$ so $$\mathrm{Ann}(xz) + \mathrm{Ann}(y) \subset \mathfrak{p} + \mathfrak{p} = \mathfrak{p}$$ But from (iii) we know that $$\mathrm{Ann}(xz) + \mathrm{Ann}(y) = R$$ which is the desired contradiction.