Let $R$ be a UFD and $K$ be its field of fractions. Let $A$ be an ideal of $R$. Define for this $A$, the $R$-submodule $A^{-1}$ of $K$ given by $$A^{-1}=\{ \alpha\in K \,\,:\,\, \alpha A\subseteq R\}.$$
Now assume that ideal $A$ is finitely generated, say for simplicity by two elements $x_1,x_2$.
Let $d=\gcd(x_1,x_2)$. Then clearly $\frac{1}{d}\in A^{-1}$.
Q. Is it true that $A^{-1}$ is $R$-module generated by single element $\frac{1}{d}$, where $d=\gcd(x_1,x_2)$?
[If this has a positive answer, then two-generated ideal can be replaced by any finitely generated ideal. I am curious about whether the inverses of finitely generated ideals of UFD are one-generated $R$-modules?]
Suppose $R$ is a UFD, with $K$ as its field of fractions.
Let $x_1,x_2\in R$, not both zero, let $d=\gcd(x_1,x_2)$, and let $A$ be the ideal $(x_1,x_2)$.
Let $D$ be the $R$-submodule of $K$ generated by ${\large{\frac{1}{d}}}$.
Claim:$\;A^{-1}=D$.
The proof is straightforward . . .
As you noted, ${\large{\frac{1}{d}}}\in A^{-1}$, hence $D\subseteq A^{-1}$.
To prove the reverse inclusion, let ${\large{\frac{p}{q}}}\in A^{-1}$, with $\gcd(p,q)=1$.
We want to show ${\large{\frac{p}{q}}}\in D$.
Since ${\large{\frac{p}{q}\in A^{-1}}}$, we have ${\large{\frac{p}{q}}}\cdot x_1\in R$, and ${\large{\frac{p}{q}}}\cdot x_2\in R$, hence $q{\,|\,}x_1$ and $q{\,|\,}x_2$.
Thus, $q$ is a common divisor of $x_1,x_2$, hence $q{\,|\,}d$.
Writing $d=eq$, for some $e\in R$, we get $$\frac{p}{q}=\frac{ep}{eq}=\frac{ep}{d}=(ep)\cdot \frac{1}{d}\in D$$ as was to be shown.