For a scheme $X$, let $LE(X)$ denote the lisse-etale site on $X$. This is the full subcategory of $\textbf{Sch}/X$ consisting of smooth morphisms to $X$, equipped with the etale topology. Let $\mathcal{O}\_X$ denote the presheaf on $LE(X)$ sending
$$(U\rightarrow X)\mapsto \Gamma(U,\mathcal{O}\_U)$$
This is in fact represented by $\mathbb{A}^1\_X$, and hence is a sheaf (of sets)


Now let $k$ be an alg. closed field. Let $X := \mathbb{A}^1\_k$ with coordinate $t$. Let $pt := \text{Spec }k$.


Consider the map $i : pt\rightarrow X$ sending $pt\mapsto (t=0)$. By Yoneda, it is possible to show that $i^\*\mathcal{O}\_X = \mathcal{O}\_{pt}$ (ie, the inverse image sheaf of the functor of points of $\mathbb{A}^1\_X$ in $LE(X)$ is the functor of points of $\mathbb{A}^1\_{pt}$ in $LE(pt)$).


In particular, $i^\*\mathcal{O}\_X(pt) = k$. **I don't see how this agrees with the definition of $i^\*$.**


By definition, $i^\*\mathcal{O}\_X$ is the sheafification of the presheaf defined as follows:


For every $Z\in LE(pt)$, consider the category $I\_Z$ consisting of objects $(U,\rho)$ where $U\in LE(X)$ and $\rho : Z\rightarrow U\_{pt} := U\times\_X pt$ is a morphism in $LE(pt)$. A morphism $(U,\rho)\rightarrow (U',\rho')$ is a morphism $g : U\rightarrow U'$ in $LE(X)$ such that $g\_{pt}\circ \rho = \rho'$.


Then, we consider the presheaf on $LE(pt)$:
$$(i^\*)^{pre}\mathcal{O}\_X : (Z\rightarrow pt)\mapsto \varinjlim\_{(U,\rho)\in I\_Z^{op}} \mathcal{O}\_X(U) = \varinjlim\_{(U,\rho)\in I\_Z^{op}}\Gamma(U,\mathcal{O}\_U)$$
where given $h : Y\rightarrow Z$ in $LE(pt)$, we get a restriction map given by the functor $I\_Z\hookrightarrow I\_Y$ sending $(U,\rho)\mapsto (U,\rho\circ h)$. Then, we define $i^\*\mathcal{O}\_X$ to be the sheafification of this presheaf.


(So far this is the setup on p45-47 of Olsson's "Algebraic spaces and stacks")


My issue is: by the definition of $i^\*\mathcal{O}\_X$, there is a natural map
$$k[t] = \mathcal{O}\_X(X)\rightarrow i^\*\mathcal{O}\_X(pt) = k$$
Which can't be injective, and for example presumably sends $t,t^2$ both to 0. However, the global sections of the presheaf:
$$(i^\*)^{pre}\mathcal{O}\_X(pt) = \varinjlim\_{(U,\rho)\in I\_{pt}^{op}}\mathcal{O}\_X(U),$$
contains $\mathcal{O}\_X(X)$ as a subset (since every $U\rightarrow X$ is smooth, hence flat, hence induces injections on global sections). Thus, the images of $t,t^2$ are distinct in $(i^\*)^{pre}\mathcal{O}\_X(pt)$. This means that if $t,t^2$ both map to 0 in $i^\*\mathcal{O}\_X(pt)$, they must agree in some etale covering of $pt$, but this seems impossible since $k$ is alg. closed, so the etale covers carry no additional information.


EDIT: Here is the argument that $i^\*\mathcal{O}\_X\cong\mathcal{O}\_{pt}$. We have, for any sheaf $F$ on $LE(pt)$:
$$Hom\_{LE(pt)}(i^\*h\_{\mathbb{A}^1\_X},F) = Hom\_{LE(X)}(h\_{\mathbb{A}^1\_X},i\_\*F) = (i\_\*F)(\mathbb{A}^1\_X) := F(i^\*\mathbb{A}^1\_X) = F(\mathbb{A}^1\_{pt}) = Hom\_{LE(pt)}(h\_{\mathbb{A}^1\_{pt}},F)$$
where we've used Yoneda twice, in the second and final equalities. But then, since this holds for any sheaf $F$ on $LE(pt)$, comparing the last term with the first, by covariant Yoneda we find that $i^\*h\_{\mathbb{A}^1\_X} = h\_{\mathbb{A}^1\_{pt}}$ Since $\mathcal{O}\_X$ is represented by $\mathbb{A}^1\_X$ and $\mathcal{O}\_{pt}$ is represented by $\mathbb{A}^1\_{pt}$, this shows that $i^\*\mathcal{O}\_X \cong \mathcal{O}\_{pt}$.



Pulling back the lisse-etale structure sheaf via the inclusion of a point
**EDIT**


I just realized that my previous version of the answer was totally misleading. Here is what really is problematic in your argument:


Just because the images of $t$ and $t^2$ are distinct in $\mathcal{O}\_X(U)$ it does not mean that they map to distinct elements in $\varinjlim \mathcal{O}\_X(U)$.


Here is a concrete description of
$i^{-1} \mathcal{O}\_X\cong i^\*\mathcal{O}\_X$
in terms of the colimit construction you describe.


Given a an object $Z \to pt$ in $LE(pt)$,
the elements of $(i^\* \mathcal{O}\_X)(Z \to pt)$ are given by
commutative squares
$\require{AMScd}$
\begin{CD}
Z @>\rho>> U \\
@VVV @VV \pi V\\
pt @>>> X \\
\end{CD}
together with a section $\sigma \in \mathcal{O}\_X(U)$.
Another element described by
\begin{CD}
Z @>\rho'>> U' \\
@VVV @VV \pi' V\\
pt @>>> X \\
\end{CD}
and $\sigma' \in \mathcal{O}\_X(U')$ is considered
equivalent to the first element if there exists
a morphism $h\colon U \to U'$ sucht that the obvious
diagram commutes and such that the restriction
of $\sigma'$ along $h$ equals $\sigma$.


Now since $\mathcal{O}\_X$ is represented by $\mathbb{A}^1\_X$
in $LE(X)$, there is a universal global section $u \in \mathcal{O}\_X(\mathbb{A}^1\_X)$ such that for any global section
$\sigma \in \mathcal{O}\_X(U)$, there is a unique morphism
$\widetilde{\sigma}\colon U \to \mathbb{A}^1\_X$ over $X$ such that
$\sigma$ is the pull-back of $u$ along $U \to \mathbb{A}^1\_X$.
It follows that the element in $(i^\* \mathcal{O}\_X)(Z \to pt)$ as above has a canonical representative given by the commutative square
\begin{CD}
Z @>\widetilde{\sigma}\circ\rho>> \mathbb{A}^1\_X \\
@VVV @VVV\\
pt @>>> X \\
\end{CD}
and the universal global section $u \in \mathcal{O}\_X(\mathbb{A}^1\_X)$.
Clearly the set of such diagrams is in canonical bijection with the set of 
morphisms $Z \to \mathbb{A}^1\_{pt}$ over $pt$.


I keep my old answer below.


**END EDIT**


Your description of the lisse-étale site $LE(X)$ and the structure sheaf $\mathcal{O}\_X$ for a scheme $X$ is correct.


Given a morphism $f\colon X \to Y$ of schemes you get a functor $f\_\* \colon Sh(LE(X)) \to Sh(LE(Y))$ on the corresponding categories of sheaves together with a morphism
$\mathcal{O}\_Y \to f\_\*\mathcal{O}\_X$ of sheaves of rings. This basically follows from the fact that the categories $LE(X)$ and $LE(Y)$ both are restrictions of the big étale site, and it is very easy to see what this functor does.


The functor $f\_\*$ (considered as a functor of sheaves of sets) has a left adjoint $f^{-1}$, which you describe (and call $f^\*$). By adjunction, you get a corresponding morphism $f^{-1}\mathcal{O}\_Y \to \mathcal{O}\_X$ on $Sh(LE(X))$.
*A priori* this is a morphism of sheaves of sets, but since $f^{-1}$ preserves finite products (which is something which needs a proof), you get a ring structure on $f^{-1}\mathcal{O}\_Y$ and the morphism $f^{-1}\mathcal{O}\_Y \to \mathcal{O}\_X$ is a homomorphism of sheaves of rings.


**However, this is not an isomorphism in general.**


*EDIT: In fact it is an isomorphism when we are working on the lisse-étale site since by the argument given in the question.* 


To construct the left adjoint $f^\*$ of $f\_\*$ as a functor of sheaves of modules, you let $f^\*\mathcal{M} = \mathcal{O}\_X \otimes\_{f^{-1}\mathcal{O}\_Y} f^{-1}\mathcal{M}$ (with the tensor product taken as sheaf of rings). From this description, it is clear that $f^\*\mathcal{O}\_Y \cong \mathcal{O}\_X$.


**But this does not follow from the Yoneda lemma as you suggested.**


*EDIT: Well, it does, but not in the direct way suggested below.*


The reason is that $f\colon X \to Y$ is not an object of $LE(Y)$ in general, so the Yoneda lemma does not apply. In particular, this is the case in your example as $\iota\colon pt \to X$ is not smooth. However, if $f\colon X \to Y$ happens to be smooth, then the Yoneda argument works and $f^{-1}\mathcal{O}\_Y \cong \mathcal{O}\_X$.


