Although I wouldn't call it an inclusion functor, the answer is yes and in fact the forgetful functor $Set^{C^{op}} \to Set/C_0$ is monadic (as well as comonadic).

I think the most illuminating way to see this is to regard a presheaf as a set $F: X \to C_0$ over $C_0$, equipped with a $C$-action which is a map $C_1 \times_{C_0} F \to F$ satisfying suitable axioms. If $c \in C_0$ and $F c$ represents the fiber, then over $c$ the action is a function of the form

$$\sum_d \hom(c, d) \times F d = \sum_d \sum_{f: c \to d} F d \to F c,$$

a many-object generalization of the way that an $M$-set over a monoid $M$ is given by a map $M \times X \to X$ satisfying suitable conditions. Just as $M \times X$ is the (underlying set of the) free $M$-set on $X$, so the domain described above, or more precisely the pullback $C_1 \times_{C_0} X$ as in the diagram

$$\begin{array}{ccc}
C_1 \times_{C_0} X & \to & X \\
\downarrow & & \downarrow F \\
C_1 & \underset{d_1}{\to} & C_0,
\end{array}
$$

is the (underlying indexed set of the) free $C$-module on $F: X \to C_0$. This is a presheaf over $C$ in an evident way: the requisite $C$-action is given by

$$C_1 \times_{C_0} C_1 \times_{C_0} X \stackrel{\text{comp} \times_{C_0} X}{\longrightarrow} C_1 \times_{C_0} X$$

Summarizing: the functor given on objects by $(F: X \to C_0) \mapsto C_1 \times_{C_0} X$, with this $C$-action just described, gives the left adjoint $Set/C_0 \to Set^{C^{op}}$ to the forgetful functor.

This is covered in lots of places, for example Mac Lane and Moerdijk's Sheaves in Geometry and Logic.