Saturated set and union of prime ideals.

First of all, it is important to use correct notation. It is $A\setminus S$, not $A/S$, the complement of $S$. It is sufficient to prove that if $S$ is saturated then every element $x\in A\setminus S$ is contained in a prime ideal which is contained in $A\setminus S$. So let $x\in A\setminus S$. Let:

$\mathcal{F}=\{I\subseteq A|\ \text{I is an ideal, $x \in I$ and $I\cap S=\emptyset$}\}$

Since $S$ is saturated, the ideal $Ax=\{ax: a\in A\}$ belongs to $\mathcal{F}$, and so $\mathcal{F}$ is not empty. Using Zorn's lemma it is very easy to show that $\mathcal{F}$ contains a maximal element with respect to inclusion. So let $P\in\mathcal{F}$ be a maximal element. We will show it is prime.

Suppose $y,z\notin P$ satisfy $yz\in P$. The ideal $P+Ay$ properly contains $P$, and so by maximality it follows that $P+Ay$ intersects $S$. So there are $p_1\in P, a_1\in A$ such that $p_1+a_1y\in S$. Similarly, there are $p_2\in P, a_2\in A$ such that $p_2+a_2z\in S$. Since $S$ is multiplicative, the product $(p_1+a_1y)(p_2+a_2z)$ belongs to $S$ as well. But:

$(p_1+a_1y)(p_2+a_2z)=p_1p_2+p_2a_1y+p_1a_2z+a_1a_2yz\in P$

Which is a contradiction to $P\cap S=\emptyset$.