I'm looking for a $S$-Scheme $X$ (with structural morphism $f$) and an additive (or rather $\newcommand{\O}{\mathcal{O}}f^{-1} O_S$-linear) endomorphism of sheaves $D: \O_X \to O_X$ which is not a differential operator (as defined in EGA IV, chapter 16).
A morphism $D: \mathcal{F} \to \mathcal{G}$ for some $\O_X$-modules $\mathcal{F}$ and $\mathcal{G}$ that isn't a differential operator would be interesting, too.
To my surprise I haven't been able to come up with such a thing. The closest I have got is the isomorphism of sheaves of abelian groups $\O_\mathbb{C} \to \O_\mathbb{C} / \underline{\mathbb{C}}$ given by integration ($\O_\mathbb{C}$ being the sheaf of holomorphic functions on $\mathbb{C}$ and $\underline{\mathbb{C}}$ the constant sheaf).
However this approach doesn't seem to work for something like $\mathbb{A}^1_\mathbb{C}$ because differentiation isn't surjective there.
A quick definition of differential operators: A map of sheaves ob abelian groups $D: \mathcal{O}_X \to \mathcal{O}_X$ is a differential operators if it factors as $u \circ d^n_{X/S}$ for some $\O_X$-linear $u$ and some $n$.
$d^n_{X/S}: \O_X \to \mathcal{P}^n_{X/S}$ is the map into the sheaf of principal parts $\mathcal{P}^n$ which in turn is defined as $\Delta_{X/S}^{-1} \O_{X \times_S X} / \mathcal{I}^{n+1}$, $\mathcal{I}$ being the kernel of $\Delta^{-1}_{X/S} \O_{X \times_S X} \to \O_X$.
The projection $p_1: X \times_S X$ yields a morphism $\O_X \to \Delta^{-1}_{X/S}\O_{X \times_S X}$ with gives $\mathcal{P}^n_{X/S}$ its structure of $\O_X$-Module. The analogous morphism arising from $p_2$ is $d^n_{X/S}$.
For an more understandable definition, refer to EGA.
EDIT: I think it is not implausible to assume that there might no such "non-differential operator" (at least not between sheaves $\mathcal{F}$, $\mathcal{G}$ locally of finite type) on sufficiently nice Schemes: On a real smooth manifold, by peetre's theorem every morphism of sheaves $C^\infty \to C^\infty$ isa differential operator.
Also for $\mathcal{F} = \bigoplus_{\mathbb{N}} \O_X$ , $D = \frac{\partial}{\partial x_1} + \frac{\partial^2}{\partial {x_2}^2} + \frac{\partial^3}{\partial {x_3}^3} + \cdots : \mathcal{F} \to \O_X$ is a morphism of sheaves of abelian groups that isn't a differential operator.
For $X = \mathbb{A}^1_\mathbb{C}$ this corresponds to finding a $\mathbb{C}$-linear map $\phi: \mathbb{C}(z) \to \mathbb{C}(z)$, which isn't a differential operator, such that if $\phi(f)$ is divisible by $\frac1{z - a}$ for any $a \in \mathbb{C}$ then so is $f$ (so the possible "domain" of $\phi(f)$ is at least as large as that of $f$, inducing a map of sheaves).
Every element $f/g \in \mathbb{C}(z)$ has a unique representation $$ \frac{f}{g} = q + \frac{r}{g} $$ (unique meaning that $q$ and $r/g$ are unique) such that $q \in \mathbb{C}[z]$ and $\deg r <\deg g$. The corresponding map $\lfloor \,\cdot\, \rfloor: f/g \mapsto q$ satisfies the above conditions if we check that it is not a differential operator.
Every differential operator $D$ on $\mathbb{A}^1_\mathbb{C}$ is of the form $$ f_0 + f_1 \frac{d}{d z} + f_2 \frac{d^2}{d z^2} + \cdots + f_n \frac{d^n}{d z^n}, \quad f_i \in \mathbb{C}[z], f_n \neq 0 $$ (We say that $n$ is the degree of $D$).
Assume $\lfloor \,\cdot\, \rfloor$ is a differential operator. Since $\lfloor z^m \rfloor = z^m$ for $m \geq 0$, we obtain inductively that $f_0$ has to be $1$ and all $f_i$ have to be $0$ for $i \geq 1$ ($D(1) = 1$ implies $f_0 = 1$, then $D(z) = z$ implies $f_1 = 0$ and so on). Hence $\lfloor \,\cdot\, \rfloor = \operatorname{id}$ contradicting $\lfloor z^{-m} \rfloor = 0$ for $m \geq 0$.
EDIT: One can actually observe, that for any field $k$
\begin{equation} \{1, x, x^2, \dots\} \cup \left.\left\{\frac{x^k}{p^n} ~\right|~ p ~\text{prime in}~ k[x], n \in \mathbb{N}^+, k < \deg p \right\} \end{equation}
is a $k$-vector space basis of $k(x)$ via partial fraction decomposition. The $k$-Endomorphisms of $k(z)$ that induce a map of sheaves are those in which basis elements of the form $x^k/p^n$ are mapped to linear combinations of $\{1, x, x^2, \dots\} \cup \{x^\ell / p^m ~|~ m \in \mathbb{N}^+, \ell < \deg p\}$ and basis elements of the form $x^k$ are mapped to polynomials.