Given an integral domain $R$, let $K$ be the field of fractions of $R$. Then, it is easily seen that if $R$ is a unique factorization domain, then the quotient group $K^{\times}/R^{\times}$ is a free abelian group.
Conversely, if $K^{\times}/R^{\times}$ is a free abelian group, then must $R$ be a UFD?
Special cases:
If $K^{\times}/R^{\times}$ is the trivial group, then that means that $R$ is already a field. Fields are trivially UFDs.
If $K^{\times}/R^{\times}$ is isomorphic to $\mathbb{Z}$, then let $\frac{a}{b}R^{\times}$ be either of the two generators of that infinite cyclic group. Then, one would be done if one could show that either $a \mid b$ (so $\frac{b}{a} \in R$) or $b \mid a$ (so $\frac{a}{b} \in R$), and then $R$ would be a discrete valuation ring with uniformizer $\frac{b}{a}$ or $\frac{a}{b}$, respectively.
$\Bbb{Z}[\frac{1+\sqrt{-7}}{2}]$ is a PID thus $$\Bbb{Q}[\sqrt{-7}]^*/\Bbb{Z}[\frac{1+\sqrt{-7}}{2}]^\times=\Bbb{Q}[\sqrt{-7}]^*/\Bbb{Z}[7\sqrt{-7}]^\times$$ is a free abelian group,
but $\Bbb{Z}[7\sqrt{-7}]$ is not a UFD: $7^3 = -(7\sqrt{-7})^2$