Let $R$ be a finite ring extension of $k[x]$ such that $R$ is free of rank $n$ as a $k[x]$-module. Moreover, let $M$ be an $R$-module, which is contained in $\operatorname{Frac}(R) = R \otimes_{k[x]} k(x)$ and is also free of rank $n$ over $k[x]$, say $M = \oplus_{i=1}^n\frac{m_i}{a_i} k[x]$ with $m_i \in R$ and $a_i \in k[x]$.
Consider the same situation analoguously with $k[x^{-1}]$, $S$ finite extension of $k[x^{-1}]$ such that $S$ is free of rank, $N$ an $S$-module contained in $\operatorname{Frac}(S) = S \otimes_{k[x^{-1}]} k(x) = \operatorname{Frac}(R)$ and is also free of rank $n$ over $k[x^{-1}]$, say $N = \oplus_{i=1}^n\frac{z_i}{b_i} k[x^{-1}]$ with $z_i \in S$ and $b_i \in k[x^{-1}]$.
Now assume that $M_x = N_{x^{-1}}$, that is $$M_x = \bigoplus_{i=1}^n\frac{m_i}{a_i} k[x,x^{-1}] = \bigoplus_{i=1}^n\frac{z_i}{b_i} k[x,x^{-1}] = N_{x^{-1}}.$$
Can we conclude that $a := \prod_{i=1}^n a_i$ and $b := \prod_{i=1}^n b_i$ are associated in the ring $k[x,x^{-1}]$?
Note that the rings $R,\,S$ are of dimension one, noetherian but need not be UFD.