Translated to just $R$, this comes down to proving that if for every prime ideal $I$ in $R$, there exists some $d \in R \setminus I$ with $a | rd$, then $a|r$.
I found a proof in case R is a UFD: Write $a = u p_1^{k_1} \ldots p_n^{k_n}$ with all $p_j$ non-associated irreducible elements and $u$ a unit. Then using the previous paragraph on $I=(p_j)$ tells us that $p_j^{k_j} | r$. Therefore $a | r$.
I've yet to find anything useful in the general case. Thanks for any help!
|for divisibility, use\mid(for example). This gives better spacing.$$a \mid b$$See also TeX/116580. – Martin Brandenburg Dec 17 '24 at 22:05\mid. Arturo and I had a discussion about this long ago here. There are actually some popular number theory textbooks that use the narrower spacing. – Bill Dubuque Dec 17 '24 at 23:39