This is a problem of Tennison Sheaf Theory Problem 2.1
"Let $I=[0,1]\subset R$, Show there is a unique(up to isomorphism) sheaf of abelian groups $F$ on $I$ with stalks:$F_0=F_1=Z, F_x=\{0\}$ for $x\in I-\{0,1\}$"
This is basically constructed from patching two skyscraper sheaves. Consider $F(U)=Z$, for either $0$ or $1\in U$, and $F(U)=0$, for anything else. This gives rise to $F$.
I think the problem can be simplified to consider non-trivial stalk at only 1 point and generalized to consider any number of closed points that are not limit points of each other.
If my proof is correct, then the statement can be relaxed to the following statement which should be wrong for sure.
Given any two sheaves $S,S'$ over $X$, if for all $x\in X$, $S_x= S'_x$, then $S\cong S'$.
Let $\Gamma(-,F)$ be the functor of sheaf of sections of sheaf space $(F,p,X)$ where $p:F\to X$. Let $L$ be the functor forming sheaf space $(LP,\pi, X)$ from presheaf $P$ where $\pi:LP\to X$ by projection.
It is clear that $\Gamma(-,LS)\cong S$ if $S$ is a sheaf. Let $S'$ be another sheaf over $X$ with same stalks as those of $S$. So we know $LS=LS'$ as $LS$ is disjoint union of all stalks. So $\Gamma(-,LS)=\Gamma(-,LS')$. So I conclude that $S\cong\Gamma(-,LS)=\Gamma(-,LS')\cong S'$ where the middle is just equality. So $S\cong S'$.
It seems that I can even say $LS\cong LS'$ as sheaf space isomorphism resulting in $\Gamma(-,LS)\cong\Gamma(-,LS')$. It is known that without prescribing the morphism $S\to S'$, even if the stalks are isomorphic, we are not guaranteed that $S\cong S'$.
- I am certain I have done something very wrong here. It would be kind that someone could point out my mistake.