Let $A$ be a domain (commutative ring with unity and no zero divisors) and let $S$ be a multiplicative subset of $A$. Denote by $S^{-1}A$ the ring of quotients $\frac{a}{s}$ (you define the ring structure as you would expect). Let $\mathfrak{p}_1$ and $\mathfrak{p}_2$ be two distinct prime ideals of $A$ such that $\mathfrak{p}_1 \cap S = \emptyset$ and $\mathfrak{p}_2 \cap S = \emptyset$. Now I would like to prove that $S^{-1}\mathfrak{p}_1$ and $S^{-1}\mathfrak{p}_2$ are two distinct prime ideals of $S^{-1}A$.
Clearly $S^{-1}\mathfrak{p}_1$ and $S^{-1}\mathfrak{p}_2$ are two ideals of $S^{-1}A$. Suppose we have two elements $\frac{a_1}{s_1}$ and $\frac{a_2}{s_2}$ in $S^{-1}A$ such that $\frac{a_1a_2}{s_1s_2} \in S^{-1}\mathfrak{p}_1$. There exists an element $\frac{p}{s_3} \in S^{-1}\mathfrak{p}_1$ such that $\frac{a_1a_2}{s_1s_2} = \frac{p}{s_3}$. Can I conclude that $a_1a_2s_3=ps_1s_2$ or do I have to say that there exists an element $t \in S$ such that $ta_1a_2s_3=tps_1s_2$?
Either way we conclude that $(t)a_1a_2s_3 \in \mathfrak{p}_1$. Because $(t)s_3 \in S$ and $S\cap\mathfrak{p}_1 = \emptyset$ and $\mathfrak{p}_1$ is a prime ideal we find $a_1 \in \mathfrak{p}_1$ or $a_2 \in \mathfrak{p}_1$ and therefore $\frac{a_1}{s_1} \in S^{-1}\mathfrak{p}_1$ or $\frac{a_2}{s_2} \in S^{-1}\mathfrak{p}_1$. We conclude that $S^{-1}\mathfrak{p}_1$ is a prime ideal.
Let $S^{-1}\mathfrak{p}_1 = S^{-1}\mathfrak{p}_2$. Let $x \in \mathfrak{p}_1$, then a fortiori $x \in S^{-1}\mathfrak{p}_1=S^{-1}\mathfrak{p}_2$. Now we find $n \in \mathfrak{p}_2$ and $m \in S$ such that $x=\frac{n}{m}$. Therefore we can find a $t \in S$ such that $txm=tn$ or $txm \in \mathfrak{p}_2$. Since $tx \in S$, $S \cap \mathfrak{p}_2 = \emptyset$ and $\mathfrak{p}_2$ prime, $x\in \mathfrak{p}_2$.
I guess this concludes my proof. Is it correct and isn't there is much more elegant way of proving this?
You're certainly right up to your last step, and at that critical stage you don't need to use the additional $t\in S$ since you are assuming that $R$ is a domain -- as such, $t(x-y)=0$ if and only if $x-y=0$ for $t\neq 0$, which is your case since $0\notin S$.
It doesn't seem like your last step is entirely right, and I tried to correct the typo you introduced. You're going for the contrapositive, but directly is good too. If $x\in \mathfrak p_1\setminus \mathfrak p_2$, then I claim $\frac x1\in S^{-1}\mathfrak p_1\setminus S^{-1}\mathfrak p_2$, so indeed the ideals are still distinct. Suppose that $\frac x1\in S^{-1}\mathfrak p_2$. Then it comes from some element $\frac ps$ with $s\in S,p\in\mathfrak p_2$. But this implies that $p=sx\in\mathfrak p_2$. Since $s\notin \mathfrak p_2$, $x\in\mathfrak p_2$, a contradiction.