Representation of an element of the field of fractions of a Dedekind domain as a fraction of elements which are relatively prime to a given ideal This is a generalization of this question.
Let $A$ be a Dedekind domain.
Let $K$ be the field of fractions of $A$.
Let $I$ be a non-zero ideal of $A$.
Let $\alpha$ be a non-zero element of $K$ which is relatively prime to $I$.
That is, $(\alpha)$ and $I$ have no common prime factors.
Are there elements $\beta$, $\gamma$ of $A$ with the following properties?
(1) $\alpha = \beta/\gamma$.
(2) $\beta$ and $\gamma$ are relatively prime to $I$.
 A: If $\alpha\in A$ there is nothing to prove, and thus we can assume that $\alpha\notin A$. 
Writing $(\alpha)=J_1/J_2$, where $J_i$ are integral ideals relatively prime to $I$, then $J_2\subseteq (A:\alpha)$. This shows that the ideal $(A:\alpha)=\{a\in A:a\alpha\in A\}$ is relatively prime to $I.$ Now we have to find an element $\gamma\in (A:\alpha)$ which is relatively prime to $I$. This follows immediately from this result: there exists $\gamma\in (A:\alpha)$ and a non-zero ideal $M$ such that $(\gamma)=(A:\alpha)M$ and $M+I=A$. Since $(A:\alpha)$ and $M$ are relatively prime to $I$ it follows that $(\gamma)$ is also relatively prime to $I$.
Remark. After posting my answer the OP suggested to apply this result directly to $J_2$ finding $\gamma\in J_2$ such that $(\gamma)=J_2M$ with $M+I=A$ and then observe that $(\alpha)=J_1M/J_2M$.
A: (Ack, you made a new question just before I started updating my answer)
Since we are interested in what happens at $I$, consider its semi-localization $A_I$ of $A$ formed by inverting every element of $A$ relatively prime to $I$.
This truly does kill every prime not dividing $I$ by the Chinese remainder theorem: for every other prime there is an $a \in A$ such that $a \equiv 0 \bmod \mathfrak{p}$ but $a \equiv 1 \bmod I$.
Asking if $\alpha = \beta / \gamma$ with $\gamma$ relatively prime to $I$ is thus equivalent to asking of $\alpha \in A_I$. This is true iff the prime factorization of $\alpha$ as a fraction over $A_I$ contains no primes to a negative power.
Given such an equation, $\alpha$ is furthermore relatively prime to $I$ if and only if $\beta$ is.
The question you actually asked can be phrased as asking if $\alpha$ is a unit in $A_I$ rather than simply being an element of $A_I$.
