Let $R$ be an integral domain and $S \subset R$ a multiplicative subset, such that $0 \notin S$. Let $p$ be a prime element of $R$. If $p|s$ for some $s \in S$ then $p/1 \in (S^{-1}R)^*$.
My attempt: If I am able to prove that $1/p \in S^{-1}R$, I am done. We know that $\exists r \in R: s=pr$. That means, $r/pr \in S^{-1}R$. I see that $r/pr$ and $1/p$ are equivalent, but can I even write $1/p$, since I don't know if $p \in S$? Also I don't know where I'm supposed to use that $p$ is prime (this is just the first part of the exercise, so it is possible that it doesn't come in play here).
What you have is essentially correct... except that you can't talk about $\frac{1}{p}$ unless $p$ lies in $S$, you are correct about that. That's why the problem is phrased the way it is: to show that $\frac{p}{1}$ is a unit. You can't "cancel" in the fraction, because only fractions with denominators in $S$ are allowed in $S^{-1}R$. So what you need to do is show that they are inverses by multiplying them out and showing the result is equal, in $S^{-1}R$, to $1_{S^{-1}R}$.
Indeed, the fact that $p$ is prime is irrelevant.
Proposition. Let $R$ be a commutative ring, not necessarily a domain, let $S$ be a multiplicative subset (not necessarily containing $1$), and let $\varphi\colon R\to S^{-1}R$ be the canonical map (sending $r\in R$ to $\frac{rt}{t}$ for some $t\in S$; this is well-defined and independent of the choice of $t$). If $x\in R$ divides some element of $S$, then $\varphi(x)$ is a unit in $S^{-1}R$.
If $1\in S$, then you can take the map $\varphi$ to be $\varphi(r)=\frac{r}{1}$.
Proof. Let $s\in S$ and $y\in R$ be such that $s=xy$. We show that $\frac{y}{s}\in S^{-1}R$ is the inverse of $\varphi(x)$. Indeed, we have $\varphi(x)\frac{y}{s} = \frac{xs}{s}\frac{y}{s} = \frac{xsy}{ss}$. We prove that this is equal to $\frac{s}{s}\in S^{-1}R$, as $\frac{s}{s}= 1_{S^{-1}R}$.
To show that $\frac{xsy}{ss}=\frac{s}{s}$ we need to show that there exists $s'\in S$ such that $s'(xyss-sss)=0$. But since $xyss=(xy)ss=sss$, we have $xyss-sss=0$, so taking $s'\in S$ arbitrary we have equality. Thus, $\frac{xys}{ss}=\frac{s}{s}=1_{S^{-1}R}$. This proves that $\varphi(x)$ is a unit in $S^{-1}R$, as desired. $\Box$