As Deane Yang pointed out, in general, an elliptic operator of order $s$ maps $H^{k}\rightarrow H^{k+s}$ for functions defined on $\mathbb{R}^{n}$. The pseudo-differential operator on a compact manifold is defined by patching over local coordinates. A smooth function over the whole compact manifold is of course smooth in the local coordinates as well. So a standard partition of unity argument is suffice. Since a smooth functions' derivatives are bounded on $M$, it is trivially inside every Sobolev space. Together this gives you the desired argument. 

The argument is usually expressed in terms of wave front sets. A standard theorem in distribution theory asserts that if $P$ is an operator with Schwartz kernel $K\in D'(\mathbb{R}^{m}\times \mathbb{R}^{n})$, and $K$'s wavefront set is contained in
$$
\{\xi\not=0, \eta\not=0\}
$$
Then $P$ defines maps 
$$
P:C^{\infty}_{c}(\mathbb{R}^{n})\rightarrow C^{\infty}(\mathbb{R}^{n})
$$
(see Friedlander and Joshi, for example)

In your case the Schwartz kernel of both operators can be explicitly written down and then the above theorem can be applied. But I am not sure if this is the best way to solve the problem at here. In general, the Fourier transform of a compactly supported function is a Schwartz function, and that of a compactly supported distribution is an analytic function. So I think analyticity should follow as well. But I do not have a proof over the top of my head.