= Solution
Use the standard commutative-ring convention with identity and a nonempty multiplicative subset. An <ideal> disjoint from that subset is proper. Suppose $ab\in I$ but $a,b\notin I$. Maximality among disjoint <ideals> means that both $I+(a)$ and $I+(b)$ meet $S$. Thus $s=i+ra\in S$ and $t=j+ub\in S$ for some $i,j\in I$, $r,u\in R$. Their product belongs to $S$ by multiplicative closure, but
$$
st=ij+iub+jra+ruab\in I,
$$
a contradiction. Consequently \b[$I$ is prime]. This is <prime ideal avoiding a multiplicative subset>.
For an <integral domain>, form its <field of fractions> from pairs $(a,b)$ with $b\ne0$, identifying $(a,b)$ and $(c,d)$ when $ad=bc$. Write the class as $a/b$ and define addition and multiplication by
$$
\frac ab+\frac cd=\frac{ad+bc}{bd},\qquad
\frac ab\frac cd=\frac{ac}{bd}.
$$
The domain condition makes the equivalence relation and these operations well-defined. Every nonzero $a/b$ has inverse $b/a$. The map $a\mapsto a/1$ is an injective <ring homomorphism> because $a/1=0/1$ implies $a=0$.
For the final assertion, pass to $R/I$. The image $\overline S$ is multiplicatively closed and does not contain zero, because $S\cap I=\varnothing$. Apply the allowed existence result to obtain an <ideal> $J$ of $R/I$ maximal among those disjoint from $\overline S$. The first proof makes $J$ prime. Its inverse image $P$ in $R$ is therefore prime, contains $I$ and avoids $S$. The quotient $R/P$ is an <integral domain>; compose
$$
\boxed{R\longrightarrow R/P\longrightarrow\operatorname{Frac}(R/P).}
$$
This homomorphism kills every element of $I$. No element of $S$ maps to zero, because the first map kills precisely $P$ and the second is injective. Thus it has both required properties. Commutativity is essential to the general statement; it is the ring convention under which the printed prime-ideal argument is used.
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