Let $\mathcal{B}$ be a base for a topological space $X$. We denote by $\mathbf{PSh}_\mathcal{B}$ the category of presheaves on the base $\mathcal{B}$, and by $\mathbf{PSh}$ the category of presheaves on $X$.
Consider $i: \mathbf{PSh} \to \mathbf{PSh}_\mathcal{B}$ the inclusion functor. If $i$ were to admit a left adjoint, then I wonder if it would make sense for that left adjoint to be the extension functor $e: \mathbf{PSh}_\mathcal{B} \to \mathbf{PSh}$, sending a presheaf on a base $\mathscr{F}$ to $e\mathscr{F}$, where for each open $U$ of $X$, we have $$e\mathscr{F}(U) = \varprojlim_{B\in \mathcal{B}, B\subseteq U}\mathscr{F}(B).$$
It is not hard to find a natural transformation $\eta : 1_{\mathbf{PSh}_\mathcal{B}} \to i\circ e$, which is actually a natural isomorphism. However, I am having trouble constructing a natural transformation $\varepsilon : e\circ i \to 1_{\mathbf{PSh}}$ which would play the role of the counit of the adjunction.
Is $e$ left adjoint to $i$ or am I looking for ghosts?
The basis $\mathcal{B}$ corresponds to a sub-poset $\mathbf{B}$ of the poset of opens $\mathcal{O}(X)$ of $X$. Write $j\colon\mathbf{B}\to\mathcal{O}(X)$ for the inclusion. Your map $i\colon\mathbf{PSh}\to\mathbf{PSh}_\mathcal{B}$ is the restriction functor $j^*\colon\mathrm{Fun}(\mathcal{O}(X)^\mathrm{op},\mathsf{Set})\to\mathrm{Fun}(\mathbf{B}^\mathrm{op},\mathsf{Set})$ (it is not precise to call it an inclusion functor, since there are generally more presheaves on $X$ than on $\mathcal{B}$). By general category theory, such a functor $j^*$ has in fact both a left adjoint and a right adjoint, namely left Kan extension and right Kan extension along $j$, see e.g. this nLab page. Since $\mathsf{Set}$ is complete and cocomplete, there are also explicit formulas for the left adjoint $j_!$ and right adjoint $j_*$ of $j^*$. Both of them are a kind of extension functor. In our case, the extension functor $e$ that you propose is (naturally isomorphic to) the functor $j_*$, i.e. the right adjoint to $j^*$. This can be seen from Proposition 2.8 on the same nLab page: if you spell out what they write there, you will see that they are just saying that $j_*\mathscr{F}(U)=\mathrm{lim}_{B\in\mathcal{B},B\subseteq U}\,\mathscr{F}(B)$ for any presheaf $\mathscr{F}$ on $\mathcal{B}$ and $U\subseteq X$ open, which is the same formula you used to define $e$.