Suppose $D$ is a Dirac operator acting on sections of a bundle $E$ over a manifold $M$, and define the Sobolev spaces $H^i(E)$ via the inner products $$\langle e_1,e_2\rangle_{H^i}:=\sum_{k=0}^i\langle D^k e_1,D^k e_2\rangle_{H^0},$$ where $e_1,e_2\in C_c^\infty(E)$. Then the closure $\overline{D}$ of $D$ is a bounded operator $H^1(E)\rightarrow H^0(E)$.
When $E$ is a vector bundle, $H^i(E)$ are Hilbert spaces, hence $\overline{D}$ is guaranteed to have an adjoint $\overline{D}^*:H^0(E)\rightarrow H^1(E)$. More generally, when $E$ is a bundle of Hilbert $A$-modules over some $C^*$-algebra $A$, it is still said in the literature (see for instance part 1 of Bunke's 1995 paper on Callias-type operators) that $\overline{D}^*$ also exists, even though one knows that in general, bounded operators need not be adjointable.
It seems to me that if $s\in H^1(E)$ and $t\in H^0(E)$, we have $$\langle\overline{D}s,t\rangle_{H^0} = \langle s,\overline{D}(\overline{D}^2+1)^{-1}t\rangle_{H^1},$$ and so it looks like $\overline{D}(\overline{D}^2+1)^{-1}:H^0(E)\rightarrow H^1(E)$ would be the adjoint, except that the existence of $(\overline{D}^2+1)^{-1}$ means that $D$ is essentially self-adjoint, which is true many cases but in general needs to be proved (and which is something that people often state after declaring $\overline{D}$ to be bounded adjointable).
Question -1: Is the above expression for $\overline{D}^*$ correct when $D$ is essentially self-adjoint?
Question 0: How does one show, without knowing $D$ is essentially self-adjoint, that $\overline{D}$ is adjointable? (Is there an argument using pseudo-differential calculus, for instance?)