Let $X$ be a topological space and $\mathcal{B}$ a base of the topology (i.e. it is closed under intersection and every open set is the union of elements from $\mathcal{B}$). Any functor from $\mathcal{B}$ to, say abelian groups, extends to a sheaf in, at most, one way. I.e., if the functor is not already contradicting the sheaf condition, it will extend to a sheaf in a unique way.

Now, let us assume that $\mathcal{P}$ is a prebase. That is to say, we first have to take all the finite intersections of the subsets of $\mathcal{P}$ for it to become a base.

Assume we have a functor from $\mathcal{P}$ to, say abelian groups, and assume that it can be extended to a sheaf. (I.e., the sheaf condition is not already contradicted). Will this be unique? Or is the prebase too little information?

I think it is too little (I am fairly sure), BUT, we have ~~an~~ two additional condition(s).

(First): Assume that for all finite intersections $\bigcap\limits_{i\in I}U_i, U_i\in\mathcal{P}$, there exists a $V\in\mathcal{P}$, such that $V\subseteq \bigcap\limits_{i\in I}U_i$.

(Second): ~~In particular, (I think this follows from the additional condition, if not, this condition STILL holds),~~ we know the values at the stalks.

Is this enough to uniquely define a sheaf? At the very least, to define one in a canonical way?

My attempt was to try and cover the intersections with open subsets from the prebase and then cover their intersections etc. and continue. But I would need a strong finiteness condition for this to eventually stabilise, which I do not think is satisfied.

Any ideas? Cheers.

EDIT: The "delted" text reflects the answer by Simon Henry.

Edit 2: ** Question fully answered by Simon Henry.** (Thanks).
Edit 3: Thanks for the clarification under the answer. That too is a rather interesting construction that I was not aware of. I am not sure if it will be helpful for the immediate question, but is definitely a fact worth knowing! Cheers.