@mcd You are right, the inclusion holds true only for $m\le 0.$ We can get that $\text{Op}\Sigma^0$ is bounded on $L^p$. Then we may define $\mathcal W^{s,p}=\{u\in \mathscr S'(\mathbb R^n), \forall a\in \Sigma^s, (\text{Op} a) u\in L^p\}$ and hopefully prove that Op$(\Sigma^m)$ sends continuously $\mathcal W^{s,p}$ into $\mathcal W^{s-m,p}$.

@Saal Hardali You mean certainly the $L^p$ boundedness. It is a delicate matter since some classes of pseudo-differential operators are bounded on $L^2$ and not on $L^p$. Here it is simpler since $\Sigma^m\subset S^{2m}_{1,0}$ and an operator with symbol in the larger set is indeed bounded from $W^{s,p}$ into $W^{s-2m,p}$. Maybe I was too quick for the completeness, but I do not see why it would be different, only the Sobolev filtration is different with the definition of the spaces $\mathscr H^m$ in my answer.

@Saal Hardali Considering the harmonic Oscillator $\mathcal H=-\Delta_x+\vert x\vert^2$, you can define the scale of Hilbert spaces $\mathscr H^m$ as the temperate distributions $u$ such that $\mathcal H^m u\in L^2$ and you can prove that $\mathscr H^m$ is also the set of $u$ such that $(\text{Op}a) u$ belongs to $L^2$ for any $a\in \Sigma^m.$ Another point is to prove that $\text{Op}\Sigma^0$ is included in the bounded operators on $L^p$ for $1<p<+\infty$, but it is a consequence of the same result for $S^0$.

@teagut It works with $u$ bounded compactly supported such that $u'$ is in $L^1$, by a van der Corput method. I simplified my first answer.

@MateuszWasilewski Thanks, I have corrected this.