Lat $A$ be a set and $\underline{A}$ the associated constant Zariski sheaf on the category $Sm/S$ of schemes which are smooth over $S$ for a fixed base scheme $S$. Is $\underline{A}$ already a (constant) sheaf for the Nisnevich topology on $Sm/S$?
I ask this because constant Zariski sheaves are easier to describe, which only depend on connected components.
Yes. This is fine for every topology in which covers are collections of morphism that are open for the Zariski topology and surjective on points.
To see this, because $\underline{A}$ satisfies the sheaf condition for disjoint unions, it suffices to show for $f: Y \to X$ open and surjective on points, $\underline{A}(X) \to \underline{A}(Y) \substack{ \to \\ \to} \underline{A}( Y\times_X Y)$ is a pullback square.
To do this, it is helpful to note that $\underline{A}(X)$ is the set of disjoint $A$-indexed open covers of $X$.
Given a disjoint $A$-indexed open cover $(F_a)_{a \in A}$ of $Y$ in $\underline{A}(Y)$, look at the image $f(F_a)$ of each set in $X$. This gives an $A$-indexed open cover of $Y$. We must check that if $(F_a)_{a \in A}$ satisfies the gluing condition, the cover of $X$ is disjoint.
In other words we must check that if $x \in X$, $y_1,y_2$ lie in the fiber of $Y$ over $X$, and $y_1 \in F_{a_1}$, $y_2 \in F_{a_2}$, then $a_1=a_2$. This follows from the existence of a point in $Y \times_X Y$ that maps to $y_1$ and $y_2$, which follows from the fact that $\operatorname{Spec} \kappa(y_1) \times_{ \operatorname{Spec} \kappa(x)} \operatorname{Spec} \kappa(y_2)$ is nonempty, where $\kappa(x)$ denotes the residue field at $x$.
Please allow me to rewrite Will's fantastic answer below, which I hope to be easier to understand and for others' convenience.
Let $(\mathscr{C}, \tau)$ be a Grothendieck site, where $\mathscr{C}$ is a category of schemes and $\tau$ is a topology on $\mathscr{C}$ finer than the Zariski topology, all of whose covers are collections of morphisms that are open for the Zariski topology and surjective on points. Lat $A$ be a set and $\underline{A}$ the associated constant sheaf on $\mathscr{C}$ with the Zariski topology. Then $\underline{A}$ already a (constant) sheaf on $(\mathscr{C}, \tau)$.
Proof. Let $Y\xrightarrow{f}X$ be a $\tau$-cover on the site $\mathscr{C}$. Note that $\underline{A}(X)$ is the set of locally constant functions with values in $A$, hence can be identified with all disjoint $A$-indexed open covers of $X$ (given by mapping $v\in\underline{A}(X)$ to the family $\{v^{-1}(a)\}_{a\in A}$).
Given any $u\in\underline{A}(Y)$ with $up_1=up_2$, we want to find a (unique) locally constant function $v\in\underline{A}(X)$ with $vf=u$. Of course, we have to define $v$ by letting $v^{-1}(a)=f(u^{-1}(a))$ (note that $f$ is an open map), provided that it is well-defined, that is, $f(u^{-1}(a))\cap f(u^{-1}(b))=\varnothing, \forall a\ne b\in A$.
We show this now: Otherwise, there would exist $x\in X$ and $y_1, y_2\in f^{-1}(x)$ with $u(y_1)=a, u(y_2)=b$. By scheme theory, there exists $z\in Y\times_XY$ with $p_1(z)=y_1, p_2(z)=y_2$, thus $a=u(y_1)=up_1(z)=up_2(z)=u(y_2)=b$, a contraction.
