Let $V$ be an affine variety over an algebraically closed field $k$ and 
$D \subset V$ a Cartier divisor which is normal and has an isolated singularity at $p \in D$.  

Let $\mathcal{O}\_V^\*, \mathcal{O}\_D^\*$ be the sheaves of invertible functions on $V$ and $D$. 
Then I think that we have an exact sequence $0 \rightarrow K \rightarrow \mathcal{O}\_V^\* \rightarrow \mathcal{O}\_D^\* \rightarrow 0$. 


**Question (Edited)** Is there an affine open neighbourhood of $p \in V' \subset V$ such that $H^0(V', \mathcal{O}\_{V'}^\*) \rightarrow H^0(D', \mathcal{O}\_{D'}^\*)$ is surjective where $D':= D \cap V'$? That is, can we lift a surjection of stalks to that on some open neighbourhood?


I think the Question is reduced to the following. 


**Question'** Is the cokernel of $H^0(V, \mathcal{O}\_{V}^\*) \rightarrow H^0(D, \mathcal{O}\_{D}^\*)$ finitely generated as an abelian group? 



Lifting invertible functions on a divisor to ambient affine variety
Suppose that $\bar{f} \in H^0(D, O\_D^\*)$ and consider a corresponding $f \in H^0(V, O\_V)$ (which may or may not be invertible). 


Then for every point $x \in D \subseteq V$, we let $\bar{f}'$ denote the element in the stalk $O\_{D,x}$ and $f'$ the element in the stalk $O\_{V,x}$. Since $\bar{f}'$ is not in the maximal ideal of $O\_{D,x}$, neither is $f'$ in the maximal ideal of $O\_{V,x}$. Thus $f'$ is invertible in a neighborhood of $x \in V$. Since this holds for all points $x \in D$, the vanishing locus $V(f')$ of $f'$ is away from $D$. It follows that there exists a neighborhood of $D$ where $f'$ is a unit.


Now, I just learned from 
[THIS QUESTION](https://mathoverflow.net/questions/57352/are-units-of-rings-of-functions-on-algebraic-varieties-finitely-generated-mod-c)
(that at least in the geometric setting you are interested in) the set of units of $H^0(D, O\_D)$ is finitely generated modulo constants. Thus, choose generators ${\bar f\_1}, \dots, \bar{f\_n}$ of $H^0(D, O\_D^\*)$ modulo constants. Lifting these to $f\_i \in H^0(V, O\_V)$, we can find an open set $U \subseteq V$ containing $D$ such that the $f\_i$ are invertible in $H^0(U, O\_U)$. It follows that $H^0(U, O\_U^\*) \to H^0(D, O\_D^\*)$ is surjective.


Thus it seems we can get a slightly stronger statement than what you asked for.


**Statement:** $\text{ }$ There exists an open neighborhood $U \subseteq X$ containing $D$ such that $H^0(U, O\_U^{\*}) \to H^0(V, O\_V^{\*})$ is surjective.


**EDIT:** *Perhaps in view of the newly revised question which appeared while I was typing this (the finite generation part), this is more information than required. But perhaps it will be useful to someone.*


