= Solution
Consider all <ideals> containing $I$ and disjoint from the nonempty <multiplicative subset> $S$. This collection is nonempty because it contains $I$. The <ascending chain condition> for the <Noetherian ring> gives a maximal member $P$: otherwise repeatedly choosing a strictly larger member would give an infinite strictly ascending chain of ideals. Since $S$ is nonempty and $P$ misses it, $P\ne R$.
Suppose $ab\in P$ but $a,b\notin P$. Maximality implies that $P+(a)$ and $P+(b)$ both meet $S$, so write $s=p+ra\in S$ and $t=q+ub\in S$ with $p,q\in P$. Then
$$
st=pq+pub+qra+ruab\in P,
$$
while closure under products gives $st\in S$, a contradiction. Hence
$$
\boxed{P\supseteq I,\quad P\cap S=\varnothing,\quad P\text{ is prime}.}
$$
This proves the <prime ideal avoiding a multiplicative subset> result without needing $1\in S$. The assumptions themselves rule out $0\in S$, since every ideal contains zero.
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