Let $f$ be an integral ideal of a number field $K$ (with ring of integers $\mathcal{O}$ and let $a$ and $b$ be fractional ideals of the same. Suppose that $ab^{-1} = x\mathcal{O}$ for some $x \in K$ and $x \in 1 + fb^{-1} = \{1+r : r\in fb^{-1}\}$. I am trying to show that $x^{-1} \in 1 + fa^{-1}$ but am at a loss on how to proceed.
If $K$ is such that it's ring of integers is a PID, this problem is easier, and the statement $x \in 1 + fb^{-1}$ is equivalent to a congruence relation $\alpha \equiv \beta \;\text{mod} \delta$ where $\alpha$, $\beta$, and $\delta$ are generators for $a$, $b$, and $f$, respectively. Since congruences are symmetric, the result follows.
But generally I cannot see what should replace the congruence relation.
Observe that $x\in 1+fb^{-1}$ if and only if $x-1 \in fb^{-1}$. But this means that, as fractional ideals, $$(x-1)\mathcal O\cdot a\subseteq\ fb^{-1}a = f\cdot(x\mathcal O)$$and hence$$\left(\frac{x-1}x\right)\mathcal O\cdot a \subseteq \ f$$and $$\left(\frac{x-1}x\right)\mathcal O = \left(\frac{1-x}x\right)\mathcal O=(x^{-1}-1)\mathcal O\subseteq \ fa^{-1}.$$ So $x^{-1} -1\in fa^{-1}$ and the result follows.