Let $R$ be a commutative ring and $S \subset R$ a non empty subset with $0_R \notin S$, and such that $s_1, s_2 \in S$ implies $s_1 \cdot s_2 \in S$. Let $I$ be an ideal of $R$ with $I \cap S = \emptyset$, and such that if $J$ is an ideal which contains $I$ then either $J = I$, or $J \cap S \neq \emptyset$.
I am wondering how to show that $a \cdot b \in I \Rightarrow a \in I$ or $b \in I$.
Consider the set $$J=\{x+ra \mid x\in I, r \in R\}.$$
Prove that this is an ideal containing $I$. Then, if $a \notin I$ you have $J \cap S \neq \emptyset$.
This shows that there exists some $x \in I$ and $r \in R$ so that $$x+ra \in S$$
Same way, if $b \notin I$, there exists an $y \in I, s \in R$ so that $$y+sb \in S$$
Then $$xy+ray+sbx+rsab \in S$$
But $x,y,ab \in I$ implies
$$xy+ray+sbx+rsab \in I \cap S =\emptyset$$
