What people usually call a base of the topology is a family $P$ such that if you have a finite set $U_i \in P$ then there is a covering of $\cap U_i$ by elements of $P$.
you do not necessarily need $P$ to be stable under intersection.

This is stronger than the condition you are asking, but this is the correct condition for having this kind of property.

For a counter example under the condition you are asking, I believe essentially anythings that does not satisfies the "base" condition above would do, here is the simplest example: consider the space with four points $x,y,z,t$ and the following pre-base:

$U = \{ x ,y ,t\}$;
$V = \{ z,y,t \}$;
$W = \{ t\}$.

It satisfies your condition.
The open are these three, $\emptyset$, the whole space $X$ and $Y=U \cap V = \{ y,t\}$.

A sheaf is the data of a diagram of groups $G(W)$,$G(V)$,$G(U)$ and $G(Y)$ with maps: $G(W) \leftarrow G(Y)$ ; $G(Y) \leftarrow G(U)$ ; $G(Y) \leftarrow G(V)$

and $G(\emptyset)$ and $G(X)$ are automatically defined by the sheaf condition $G(\emptyset)=\{0\}$ and $G(X) = G(U)\times_{G(Y)} G(V)$.

A sheaf in your sense, is omiting $G(Y)$, and only have a map $G(W) \leftarrow G(U) \coprod G(V)$.

any group $G(Y)$ that factor the maps above is a possible extension as a sheaf.

Also note that the value of the stalk at $y$ is $G(Y)$, so the value of the stalk is not determined by the value of $G$ on the pre-base as you claimed.