[proofplan]
We reduce the semiclassical estimate to the ordinary $h=1$ Calderón-Vaillancourt estimate by a unitary dilation. The dilation converts $\operatorname{Op}_h(a)$ into a standard pseudodifferential operator whose symbol is $b_h(X,\Xi)=a(h^{1/2}X,h^{1/2}\Xi;h)$, and every derivative seminorm of $b_h$ is bounded by the corresponding derivative seminorm of $a$. We then invoke the standard Calderón-Vaillancourt estimate at $h=1$, which gives $L^2$ boundedness from finitely many order-zero symbol seminorms. The resulting constants depend only on dimension, $h_0$, and finitely many symbol seminorms, hence are uniform for $0<h\le h_0$.
[/proofplan]
[step:Reduce the semiclassical operator to a standard quantization by dilation]
For $h \in (0,h_0]$, define the unitary dilation $U_h: L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ by
\begin{align*}
(U_h v)(x) = h^{-n/4}v(h^{-1/2}x).
\end{align*}
Its inverse is given by
\begin{align*}
(U_h^{-1}u)(X) = h^{n/4}u(h^{1/2}X).
\end{align*}
For fixed $h$, define the rescaled symbol $b_h \in C^\infty(\mathbb{R}^n_X\times\mathbb{R}^n_\Xi)$ by
\begin{align*}
b_h(X,\Xi) = a(h^{1/2}X,h^{1/2}\Xi;h).
\end{align*}
We claim that
\begin{align*}
U_h^{-1}\operatorname{Op}_h(a)U_h = \operatorname{Op}_1(b_h),
\end{align*}
where
\begin{align*}
(\operatorname{Op}_1(b_h)v)(X) = (2\pi)^{-n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} e^{i(X-Y)\cdot \Xi}b_h(X,\Xi)v(Y)\,d\mathcal{L}^n(Y)\,d\mathcal{L}^n(\Xi).
\end{align*}
Indeed, for $v \in \mathcal{S}(\mathbb{R}^n)$ and $X \in \mathbb{R}^n$, substitute $x=h^{1/2}X$, $y=h^{1/2}Y$, and $\xi=h^{1/2}\Xi$. The measure transforms are $d\mathcal{L}^n(y)=h^{n/2}d\mathcal{L}^n(Y)$ and $d\mathcal{L}^n(\xi)=h^{n/2}d\mathcal{L}^n(\Xi)$, while
\begin{align*}
(2\pi h)^{-n}d\mathcal{L}^n(y)d\mathcal{L}^n(\xi) = (2\pi)^{-n}d\mathcal{L}^n(Y)d\mathcal{L}^n(\Xi).
\end{align*}
The phase satisfies
\begin{align*}
(x-y)\cdot \xi/h = (X-Y)\cdot \Xi.
\end{align*}
The factors $h^{n/4}$ and $h^{-n/4}$ from $U_h^{-1}$ and $U_h$ cancel, giving the asserted identity.
Moreover, for every pair of multi-indices $\alpha,\beta \in \mathbb{N}_0^n$, the chain rule gives
\begin{align*}
\partial_X^\alpha\partial_\Xi^\beta b_h(X,\Xi)=h^{(|\alpha|+|\beta|)/2}(\partial_x^\alpha\partial_\xi^\beta a)(h^{1/2}X,h^{1/2}\Xi;h).
\end{align*}
Let $N_0=N_0(n)$ denote the finite derivative order in the standard Calderón-Vaillancourt estimate invoked in the next step. Then for every $|\alpha|+|\beta|\le N_0$ and every $0<h\le h_0$,
\begin{align*}
\sup_{(X,\Xi)}|\partial_X^\alpha\partial_\Xi^\beta b_h(X,\Xi)|\le \max\{1,h_0^{N_0/2}\}\sup_{(x,\xi)}|\partial_x^\alpha\partial_\xi^\beta a(x,\xi;h)|.
\end{align*}
Thus the rescaling introduces the explicit factor $\max\{1,h_0^{N_0/2}\}$, so it remains to prove the $h=1$ estimate with constants depending only on $n$ and finitely many derivatives of the symbol.
[guided]
The semiclassical parameter appears both in the prefactor $(2\pi h)^{-n}$ and in the phase $e^{i(x-y)\cdot \xi/h}$. The natural way to remove it is to scale position and frequency by $h^{1/2}$ in opposite places inside the oscillatory integral. We use the unitary map $U_h: L^2(\mathbb{R}^n)\to L^2(\mathbb{R}^n)$,
\begin{align*}
(U_h v)(x)=h^{-n/4}v(h^{-1/2}x).
\end{align*}
The power $h^{-n/4}$ is chosen exactly so that
\begin{align*}
\|U_h v\|_{L^2(\mathbb{R}^n)}^2 = \int_{\mathbb{R}^n}h^{-n/2}|v(h^{-1/2}x)|^2\,d\mathcal{L}^n(x)=\|v\|_{L^2(\mathbb{R}^n)}^2
\end{align*}
after the change of variables $x=h^{1/2}X$.
Now define $b_h:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{C}$ by
\begin{align*}
b_h(X,\Xi)=a(h^{1/2}X,h^{1/2}\Xi;h).
\end{align*}
We compute $U_h^{-1}\operatorname{Op}_h(a)U_h$ on a Schwartz function $v$. In the kernel formula for $\operatorname{Op}_h(a)$, set $x=h^{1/2}X$, $y=h^{1/2}Y$, and $\xi=h^{1/2}\Xi$. Then
\begin{align*}
d\mathcal{L}^n(y)=h^{n/2}d\mathcal{L}^n(Y)
\end{align*}
and
\begin{align*}
d\mathcal{L}^n(\xi)=h^{n/2}d\mathcal{L}^n(\Xi).
\end{align*}
Therefore the semiclassical normalization becomes the standard normalization:
\begin{align*}
(2\pi h)^{-n}d\mathcal{L}^n(y)d\mathcal{L}^n(\xi)=(2\pi)^{-n}d\mathcal{L}^n(Y)d\mathcal{L}^n(\Xi).
\end{align*}
The phase also becomes standard:
\begin{align*}
\frac{(x-y)\cdot\xi}{h}=(X-Y)\cdot\Xi.
\end{align*}
The two dilation factors $h^{n/4}$ and $h^{-n/4}$ cancel. Hence
\begin{align*}
U_h^{-1}\operatorname{Op}_h(a)U_h=\operatorname{Op}_1(b_h).
\end{align*}
It remains to check that the rescaled symbol has no worse seminorms. By the chain rule,
\begin{align*}
\partial_X^\alpha\partial_\Xi^\beta b_h(X,\Xi)=h^{(|\alpha|+|\beta|)/2}(\partial_x^\alpha\partial_\xi^\beta a)(h^{1/2}X,h^{1/2}\Xi;h).
\end{align*}
For the finite set of derivatives used in the Calderón-Vaillancourt estimate, say $|\alpha|+|\beta|\le N_0$, the factor $h^{(|\alpha|+|\beta|)/2}$ is bounded on $0<h\le h_0$ by $\max\{1,h_0^{N_0/2}\}$. Therefore the $h=1$ theorem applied to $b_h$ gives the desired semiclassical estimate for $a$ after conjugating back by the unitary $U_h$, with the final constant depending on $n$ and $h_0$.
[/guided]
[/step]
[step:Invoke the standard Calderón-Vaillancourt estimate at unit semiclassical parameter]
Let $b\in C^\infty(\mathbb{R}^n_X\times\mathbb{R}^n_\Xi)$ and define $B:=\operatorname{Op}_1(b)$ on $\mathcal{S}(\mathbb{R}^n)$ by
\begin{align*}
(Bv)(X)=(2\pi)^{-n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}e^{i(X-Y)\cdot\Xi}b(X,\Xi)v(Y)\,d\mathcal{L}^n(Y)\,d\mathcal{L}^n(\Xi).
\end{align*}
We use the standard Calderón-Vaillancourt theorem for Kohn-Nirenberg quantization at parameter $1$. Its hypotheses are satisfied because $b$ is smooth on $\mathbb{R}^n\times\mathbb{R}^n$ and the seminorms appearing below are finite. Therefore there are $N_0=N_0(n)$ and $C_0=C_0(n)>0$ such that
\begin{align*}
\|Bv\|_{L^2(\mathbb{R}^n)}\le C_0\left(\sum_{|\alpha|+|\beta|\le N_0}\sup_{(X,\Xi)}|\partial_X^\alpha\partial_\Xi^\beta b(X,\Xi)|\right)\|v\|_{L^2(\mathbb{R}^n)}.
\end{align*}
[guided]
The purpose of this step is not to reprove Calderón-Vaillancourt; it is to isolate exactly which standard result is needed. The standard Calderón-Vaillancourt theorem applies to the Kohn-Nirenberg operator
\begin{align*}
(Bv)(X)=(2\pi)^{-n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}e^{i(X-Y)\cdot\Xi}b(X,\Xi)v(Y)\,d\mathcal{L}^n(Y)\,d\mathcal{L}^n(\Xi)
\end{align*}
when $b$ is smooth and has bounded derivatives up to a finite order depending only on the dimension. Those are precisely the hypotheses recorded here: $b\in C^\infty(\mathbb{R}^n\times\mathbb{R}^n)$, and the finite seminorm sum
\begin{align*}
\sum_{|\alpha|+|\beta|\le N_0}\sup_{(X,\Xi)}|\partial_X^\alpha\partial_\Xi^\beta b(X,\Xi)|
\end{align*}
is finite. The theorem therefore gives
\begin{align*}
\|Bv\|_{L^2(\mathbb{R}^n)}\le C_0\left(\sum_{|\alpha|+|\beta|\le N_0}\sup_{(X,\Xi)}|\partial_X^\alpha\partial_\Xi^\beta b(X,\Xi)|\right)\|v\|_{L^2(\mathbb{R}^n)}.
\end{align*}
Here $N_0=N_0(n)$ and $C_0=C_0(n)>0$ are the derivative order and operator constant supplied by the standard theorem. This avoids treating oscillatory product kernels as absolutely convergent integrals and leaves the almost-orthogonality argument inside the cited Calderón-Vaillancourt result.
[/guided]
[/step]
[step:Transfer the standard estimate back to the semiclassical operator]
Apply the standard estimate to $B=\operatorname{Op}_1(b_h)$. For $N_0=N_0(n)$,
\begin{align*}
\|\operatorname{Op}_1(b_h)v\|_{L^2(\mathbb{R}^n)}\le C_0\left(\sum_{|\alpha|+|\beta|\le N_0}\sup_{(X,\Xi)}|\partial_X^\alpha\partial_\Xi^\beta b_h(X,\Xi)|\right)\|v\|_{L^2(\mathbb{R}^n)}.
\end{align*}
Using the derivative computation from the dilation step, the finite sum is bounded by
\begin{align*}
C_1\sum_{|\alpha|+|\beta|\le N_0}\sup_{(x,\xi)}|\partial_x^\alpha\partial_\xi^\beta a(x,\xi;h)|,
\end{align*}
where $C_1=\max\{1,h_0^{N_0/2}\}>0$ is independent of $h$ and $a$.
Now let $u\in\mathcal{S}(\mathbb{R}^n)$ and set $v=U_h^{-1}u\in\mathcal{S}(\mathbb{R}^n)$. Since $U_h$ is unitary and $U_h^{-1}\operatorname{Op}_h(a)U_h=\operatorname{Op}_1(b_h)$,
\begin{align*}
\|\operatorname{Op}_h(a)u\|_{L^2(\mathbb{R}^n)}=\|\operatorname{Op}_1(b_h)v\|_{L^2(\mathbb{R}^n)}.
\end{align*}
Therefore
\begin{align*}
\|\operatorname{Op}_h(a)u\|_{L^2(\mathbb{R}^n)}\le C\left(\sum_{|\alpha|+|\beta|\le N}\sup_{(x,\xi)}|\partial_x^\alpha\partial_\xi^\beta a(x,\xi;h)|\right)\|u\|_{L^2(\mathbb{R}^n)},
\end{align*}
with $N=N_0(n)$ and $C=C(n,h_0)$. This proves the asserted uniform $L^2$ boundedness of $\operatorname{Op}_h(a)$ on the interval $0<h\le h_0$.
[/step]
