I think that if we consider the category perfect modules(which we have to in order to get a non-trivial answer for Hochschild homology) then the inclusion of the subcategory generated by $E$ doesn't have any adjoints. For example, the usual construction of the left adjoint is $M\mapsto RHom(M, E)^{*}\otimes_k E$ but for $M=\mathcal{D}$(free module of rank $1$) we have $$RHom(\mathcal{D},\mathcal{O})=Hom(\mathcal{D},{\mathcal{O}})=\mathcal{O}$$ but $\mathcal{O}$ is not a finite-dimensional vector space. One could probably turn this observation into a rigorous argument by showing that the adjoint, if exists, should be compatible with that on the category of all modules.

Another way to see that there is no orthogonal decomposition of the form $D=\langle \langle \mathcal{O}\rangle, A\rangle$ is that the right(and also left) orthogonal $\mathcal{O}^{\perp}$ is zero in the category of perfect D-modules. Indeed, if $M$ is an object of the derived category of D-modules with finitely generated cohomology in a bounded range of degrees which is right orthogonal to $\mathcal{O}$, the spectral sequence $$E^{pq}_2=Ext^q(\mathcal{O},H^p(M))\Rightarrow H^{p+q}(RHom(\mathcal{O},M))$$ converges to $0$. Let $p$ be the minimal degree where $H^p(M)\neq 0$. There are no differentials coming in or out of the object $E^{p,0}_r$ for any $r$ so $E^{p,0}_2$ has to vanish for the spectral sequence to be converging to $0$. Since $\mathcal{O}$ is the cokernel of the right multiplication by $\partial_x$ on $\mathcal{D}$, it implies that $\partial_x$ is an automorphism of $H^p(M)$. But since $H^p(M)$ is finitely generated, it has to be $0$ which contradicts with the definition of $p$.