Let $X$ be a topological space, $\mathcal{B}$ a basis of $X$ and $\mathcal{F}$ be a presheaf of rings on $X$. It's well known that if $\mathcal{F}$ is a $\mathcal{B}$-sheaf then, up to isomorphisms, we have an unique shaf $\overline{\mathcal{F}}$ which extends $\mathcal{F}$.
Is it also true that $\overline{\mathcal{F}}(U)\cong\mathcal{F}(U)$ for all the open sets $U\subseteq X$?
No: here is a counterexample.
Take $X=\mathbb R$, $\mathcal F= \mathbb Z^{\operatorname {presh}}$ the constant presheaf with values in $\mathbb Z$ and let $\mathcal B$ be the set of bounded open intervals $(a,b)\subset \mathbb R$.
Then $\mathcal F$ is a sheaf on $\mathcal B$ and the associated sheaf $\widetilde {\mathcal F}= \mathbb Z^{\operatorname {sh}}$ is the sheaf of locally constant functions with values in $\mathbb Z$.
However, for $U=\mathbb R\setminus \{0\}$ we have $$\widetilde {\mathcal F}(U)=\mathbb Z^2 \not\cong \mathcal F(U)=\mathbb Z$$
Edit
Note however that (not only here but in the general situation of the question) for $B\in \mathcal B$ we do have $\widetilde {\mathcal F}(B)= \mathcal F(B)$