As discussed in the comments, the statement probably needs to be modified in order for $\langle D \rangle^{-n}$ to be defined. I'm guessing that the correct statement should fit into the following framework:

Proposition: Let $D$ be an essentially self-adjoint first order elliptic operator on a possibly non-compact manifold $M$, let $f \in C_c^\infty(M)$, and let $g \in C_0(\mathbb{R})$. Then $m(f) g(D)$ is compact where $m(f)$ is the multiplication operator by $f$.

I'm not going to be able to drudge up all of the gory details, but in the end the proof can be pieced together using standard elliptic analysis. First consider the resolvent function $g(t) = (i + t)^{-1}$. Let $K$ denote the support of $f$ and let $v$ be a vector in the domain of the closure $\overline{D}$ of $D$, so that $v' = m(f)v$ is in the Sobolev space $L_1^2(K)$. Garding's inequality estimates the Sobolev $1$-norm of $v'$ in terms of the $L^2$ norm of $v'$ and of $\overline{D}v'$; from this it follows that $m(f)(i + \overline{D})^{-1}$ maps $L^2(M)$ continuously into $L_1^2(K)$. But the Rellich lemma asserts that the inclusion of $L_1^2(K)$ into $L^2(M)$ is compact, so the result is proved for the specific $g$ above. Now, the set of all $g$ for which the result is true is closed under linear combinations, pointwise multiplication, complex conjugation, and uniform limits, so by the Stone-Weierstrass theorem the result is true for any $g \in C_0(\mathbb{R})$. Notice, however, that the proposition does not apply to $g(t) = t^{-n}$, hence my concerns in the comments.

Now, the Dirac operator on a complete Riemannian manifold is essentially self adjoint and therefore fits into the proposition above. This ultimately follows from the fact that its symbol is the Clifford multiplication endomorphism which is not only invertible (away from the $0$ section) but bounded in norm on the unit cosphere bundle. In other words, it has "finite propagation speed". I'm not quite sure how this works out in the semi-Riemannian case, but my guess is that it does as long as you have some counterpart of the completeness assumption: note that the Dirac operator on $(0,1)$ is not essentially self-adjoint.