For any domain $A$ let $A^\times$ be its group of units.
Let $A$ be a noetherian domain with only finitely many prime ideals, and field of fractions $K$.
Is the group $K^\times/A^\times$ finitely generated?
The answer is no in general.
First by an answer to the question Does every Noetherian domain have finitely many height 1 prime ideals?, we know that $A$ has dimension at most $1$. Let $m_1,\dots, m_n$ be the maximal ideals of $A$. It is easy to see that inside $K^\times$, we have $A^\times =\cap_i (A_{m_i})^\times$, so the canonical map
$$ K^\times /A^\times \to \prod_i K^\times/(A_{m_i})^\times$$
is injective and we are reduced to the case where $A$ is local of dimension $\le 1$. If $A$ is integrally closed, then $K^\times /A^\times$ is generated by an uniformizing element of $A$.
So the first conclusion is $K^\times /A^\times$ is finitely generated if $A$ is integrally closed.
Now the question is what happens if $A$ is local, but not necessarily integrally closed. Let us consider an example. Let $F$ be a field and $c\in F$ an element which is not a square in $F$. Let $A$ be $F[x,y]/(y^2-cx^2)$ localized at the maximal ideal generated by $x,y$. It is integral, noetherian, with only two prime ideals.
Let $L=F[\sqrt{c}]$. It can be identified to a subring of $K=\mathrm{Frac}(A)$ by sending $\sqrt{c}$ to $y/x$. It is easy to see that $L\cap A=F$. So
$$ L^\times/F^\times \hookrightarrow K^\times/A^\times.$$
We have $L^\times=F^\times+ \sqrt{c}F^\times$, and, set-theoretically, $L^\times/F^\times$ is in bijection with $F^\times$. If we choose a uncountable field $F$, then $K^\times/A^\times$ is uncountable, hence not finitely generated. Concretly, we can take $F=\mathbb C(t)$ and $c=t$.
In this counterexample, if $B$ denotes the integral closure of $A$ in $K$, then $L\subset B$ and $B^\times/A^\times$ is not finitely generated.
The article
D. D. Anderson, “Integral Domains with Finitely Generated Groups of Divisibility”, Proceeding AMS 112 (3), 1991
most likely provides the information you seek for.
In particular theorem 0 in this article seems to show that in general the answer to your question is no: a necessary requirement seems to be that the integral closure $\overline{A}$ of $A$ in its field of fractions is a finite $A$-module, and is a Bezout domain. So e.g. if $\overline{A}$ is noetherian, it is a principal ideal domain.
Added after QiL’s post:
Putting the information from the post of QiL and the cited article together, we get the following nice result:
Theorem: Let $A$ be a noetherian domain with finite spectrum and field of fractions $K$. Let $\overline{A}$ be the integral closure of $A$ in $K$ and let $f$ be the conductor ideal of the extension $A\subseteq\overline{A}$. Then the following statements are equivalent:
$K^\times/A^\times$ is finitely generated.
For every prime ideal $p\supseteq f$ the fraction field of $A/p$ is finite.
Proof. $1\Rightarrow 2$: By theorem 1 in Anderson’s paper the ring $\overline{A}/f$ is finite. In particular the fraction fields of $\overline{A}/q$ are finite for all $f\subseteq q$. These fields contain the fraction fields of $A/p$, $p\supseteq f$.
2$\Rightarrow 1$: By QiL’s post and Theorem 3 in Andersons’s paper it suffices to prove that $\overline{A}/f$ is finite, which is equivalent to the finiteness of the fields $\overline{A}/q$, $f\subseteq q$. By Krull-Akizuki the extensions
$A/q\cap A\subseteq \overline{A}/q$ are finite. Hence the assertion follows.