This question is related to another question I asked.
However, I think this question is more general.
If $R$ is an integral domain and $a/b\in \operatorname{Frac}(R)$. Is $R[a/b]$ a UFD?
Is this statement true?
I think this is what is going on in my other question, but I can't see why it is true?
The statement is not true. Let $R$ be the set of polynomials with integer coefficients that do not have a linear term. This is an integral domain (subring of the integral domain $\mathbb{Z}[x]$), but is not a UFD, since $x^6$ can be factored as $x^3x^3$ and as $x^2x^2x^2$, with $x^2$ and $x^3$ irreducibles. Now adjoin $\frac{1}{2}$; the resulting ring is the ring of polynomials with coefficients in $\mathbb{Z}[\frac{1}{2}]$ that have no linear terms. That’s still not a UFD, for the exact same reason.
On the other hand, if $R$ is a UFD itself, then the answer is “yes”: all you are doing is localizing at the irreducible divisors of $b$, and the localization of a UFD is still a UFD.