Let me answer your first question:

No, in general not: Assume $D$ is an elliptic differential operator on $\mathcal C^\infty(M)$ satisfying the Leibniz rule $$D(fg)=f D(g)+D(f)g$$ for all $f,g\in\mathcal C^\infty(M).$ Then, $D$ must be a first order differential operator as the commutator with the multiplication operator is a zeroth order differential operator. Hence, $D= X+m$ where $X\in\mathcal X(M)$ and $m$ denotes the multiplication with $m\in\mathcal C^\infty(M)$, see for example Chapter 2 in the book Global Calculus by Ramanan. Thus, ellipticity of $D$ implies that $M$ must be 1-dimensional. Of course, any 1-dimensional manifold admits an elliptic operator on the space of smooth function which satisfies the Leibniz rule.