= Solution
By the <first isomorphism theorem for rings>, the given surjection identifies $S$ with $R/\ker\phi$. Since $R$ is a <principal ideal domain>, write $\ker\phi=(a)$.
If $a=0$, the <quotient ring> is $R$, an <integral domain>, so part (a)(ii) applies. If $a$ is a <unit>, the <quotient ring> is the <zero ring>; its unique element is associate to itself by multiplication by its identity, so it too is an <associator ring>.
In the remaining case, <unique factorization> gives
$$
a=\varepsilon p_1^{k_1}\cdots p_r^{k_r},
$$
where $\varepsilon$ is a <unit>, the $p_i$ are pairwise nonassociate <prime elements>, and $k_i\geq1$. The <principal ideals> $(p_i^{k_i})$ are pairwise comaximal: different prime factors have <greatest common divisor> $1$, and <Bézout's identity> gives a linear combination equal to $1$. The <Chinese remainder theorem> therefore gives a <ring isomorphism>
$$
R/(a)\ \cong\ \prod_{i=1}^r R/(p_i^{k_i}).
$$
Every factor is an <associator ring> by part (c), and their finite <direct product of rings> is an <associator ring> by part (b). A <ring isomorphism> preserves <units> and <principal ideals>, so the property transfers to $S$. Hence
$$
\boxed{S\text{ is an associator ring}.}
$$
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