[proofplan]
We use the Kohn-Nirenberg composition formula for properly supported scalar pseudodifferential operators to write the ordered products in the same local quantization. Namely, $AB = \operatorname{Op}(c)$ and $BA = \operatorname{Op}(d)$, where $c: U \times \mathbb{R}^n \to \mathbb{C}$ and $d: U \times \mathbb{R}^n \to \mathbb{C}$ are the Kohn-Nirenberg composition symbols of the ordered pairs $(a,b)$ and $(b,a)$. The scalar order $m+m'$ terms cancel because ordinary multiplication of scalar symbols is commutative. The first possible nonzero contribution is therefore the $|\alpha|=1$ part of the composition expansion, and modulo $S^{m+m'-2}$ it depends only on the principal symbol representatives $a_m$ and $b_{m'}$. Rewriting this first-order difference gives exactly the displayed Poisson-bracket representative, while all higher composition terms and all lower-order symbol components lie in $S^{m+m'-2}$.
[/proofplan]
[step:Apply the composition formula to the two ordered products]
Because $A$ and $B$ are properly supported, both compositions $AB$ and $BA$ are defined as properly supported pseudodifferential operators on $U$. The symbols $a \in S^m(U \times \mathbb{R}^n)$ and $b \in S^{m'}(U \times \mathbb{R}^n)$ are scalar symbols written in the same local Kohn-Nirenberg quantization convention, which are exactly the symbol-class and quantization hypotheses needed for the Kohn-Nirenberg pseudodifferential operator composition formula. We use the version of the composition formula asserting that, for properly supported scalar operators in a common Kohn-Nirenberg quantization, the ordered product has a Kohn-Nirenberg composition symbol with the displayed asymptotic expansion and with the remainder after truncating at $|\alpha|<N$ lying in $S^{m+m'-N}(U \times \mathbb{R}^n)$ for each $N \in \mathbb{N}$. Applying that composition formula to the ordered pairs $(A,B)$ and $(B,A)$, there are Kohn-Nirenberg composition symbols
\begin{align*}
c: U \times \mathbb{R}^n \to \mathbb{C}
\end{align*}
and
\begin{align*}
d: U \times \mathbb{R}^n \to \mathbb{C}
\end{align*}
for the ordered pairs $(a,b)$ and $(b,a)$, respectively, with $c \in S^{m+m'}(U \times \mathbb{R}^n)$, $d \in S^{m+m'}(U \times \mathbb{R}^n)$,
\begin{align*}
AB = \operatorname{Op}(c)
\end{align*}
and
\begin{align*}
BA = \operatorname{Op}(d).
\end{align*}
and asymptotic expansions
\begin{align*}
c \sim \sum_{\alpha \in \mathbb{N}_0^n} \frac{1}{\alpha!}\left(\frac{1}{i}\right)^{|\alpha|}(\partial_\xi^\alpha a)(\partial_x^\alpha b)
\end{align*}
and
\begin{align*}
d \sim \sum_{\alpha \in \mathbb{N}_0^n} \frac{1}{\alpha!}\left(\frac{1}{i}\right)^{|\alpha|}(\partial_\xi^\alpha b)(\partial_x^\alpha a).
\end{align*}
Here $\alpha=(\alpha_1,\dots,\alpha_n) \in \mathbb{N}_0^n$ is a multi-index,
$\alpha! := \alpha_1!\cdots \alpha_n!$, $|\alpha| := \alpha_1+\cdots+\alpha_n$,
\begin{align*}
\partial_\xi^\alpha := \partial_{\xi_1}^{\alpha_1}\cdots \partial_{\xi_n}^{\alpha_n},
\end{align*}
and
\begin{align*}
\partial_x^\alpha := \partial_{x_1}^{\alpha_1}\cdots \partial_{x_n}^{\alpha_n}.
\end{align*}
Therefore
\begin{align*}
[A,B] = \operatorname{Op}(c-d).
\end{align*}
[/step]
[step:Cancel the scalar top-order term]
The $\alpha=0$ contribution to $c$ is $ab$, while the $\alpha=0$ contribution to $d$ is $ba$. Since the symbols are scalar-valued complex functions on $U \times \mathbb{R}^n$, multiplication is commutative pointwise, so
\begin{align*}
ab - ba = 0.
\end{align*}
Thus the entire order $m+m'$ part of $c-d$ vanishes.
The composition formula also gives the order of the remaining symbol after the top-order cancellation. Truncating both asymptotic expansions after the $|\alpha|=0$ term leaves remainders in $S^{m+m'-1}(U \times \mathbb{R}^n)$. The $|\alpha|=0$ parts cancel because $ab-ba=0$, so the whole difference $c-d$ is the difference of two such remainders modulo the cancelled top-order term. Therefore
\begin{align*}
c-d \in S^{m+m'-1}(U \times \mathbb{R}^n).
\end{align*}
Consequently
\begin{align*}
[A,B] \in \Psi^{m+m'-1}(U).
\end{align*}
[/step]
[step:Compute the first surviving symbol term modulo lower order]
Let $e_j \in \mathbb{N}_0^n$ denote the multi-index with $1$ in the $j$-th component and $0$ in every other component. The $|\alpha|=1$ part of the symbol difference $c-d$ is
\begin{align*}
r_{m+m'-1} := \frac{1}{i}\sum_{j=1}^n \left(\partial_{\xi_j}a\,\partial_{x_j}b - \partial_{\xi_j}b\,\partial_{x_j}a\right).
\end{align*}
To justify the congruence modulo $S^{m+m'-2}$, we use the composition formula with truncation order $N=2$. Its remainder statement gives, after subtracting the $|\alpha|=0$ and $|\alpha|=1$ contributions from $c$ and from $d$, remainders lying in $S^{m+m'-2}(U \times \mathbb{R}^n)$. Since the $|\alpha|=0$ contributions have already cancelled, the class of $c-d$ modulo $S^{m+m'-2}$ is exactly the $|\alpha|=1$ difference. Therefore
\begin{align*}
c-d \equiv r_{m+m'-1} \pmod{S^{m+m'-2}}.
\end{align*}
[guided]
The purpose of this step is to isolate the first part of the composition expansion that can affect the principal symbol of the commutator. We are allowed to use the Kohn-Nirenberg composition formula because $A=\operatorname{Op}(a)$ and $B=\operatorname{Op}(b)$ are properly supported scalar pseudodifferential operators, $a \in S^m(U \times \mathbb{R}^n)$ and $b \in S^{m'}(U \times \mathbb{R}^n)$, and both operators are written in the same local Kohn-Nirenberg quantization. The formula gives composition symbols $c,d:U \times \mathbb{R}^n \to \mathbb{C}$ for $AB$ and $BA$, respectively, and its remainder statement says that after truncating at $|\alpha|<N$ the error lies in $S^{m+m'-N}(U \times \mathbb{R}^n)$.
For $N=2$, the expansion for $c$ modulo $S^{m+m'-2}(U \times \mathbb{R}^n)$ consists of the $\alpha=0$ term and the $|\alpha|=1$ terms. The $\alpha=0$ term is $ab$. The expansion for $d$ modulo the same lower-order class has $\alpha=0$ term $ba$. Since $a$ and $b$ are scalar-valued complex functions on $U \times \mathbb{R}^n$, pointwise multiplication is commutative, so
\begin{align*}
ab-ba=0.
\end{align*}
Thus the top-order contribution to $c-d$ vanishes, and the first possible surviving contribution comes from the multi-indices with exactly one $\xi$-derivative.
For each $j \in \{1,\dots,n\}$, let $e_j \in \mathbb{N}_0^n$ be the multi-index whose $j$-th component is $1$ and whose other components are $0$. The $e_j$ contribution to the Kohn-Nirenberg composition symbol $c$ of $AB$ is
\begin{align*}
\frac{1}{i}\partial_{\xi_j}a\,\partial_{x_j}b.
\end{align*}
The corresponding $e_j$ contribution to the Kohn-Nirenberg composition symbol $d$ of $BA$ is
\begin{align*}
\frac{1}{i}\partial_{\xi_j}b\,\partial_{x_j}a.
\end{align*}
Subtracting the second expression from the first and summing over $j$ gives the complete $|\alpha|=1$ contribution
\begin{align*}
r_{m+m'-1} := \frac{1}{i}\sum_{j=1}^n \left(\partial_{\xi_j}a\,\partial_{x_j}b - \partial_{\xi_j}b\,\partial_{x_j}a\right).
\end{align*}
The remainder statement with $N=2$ is the point that controls all terms not displayed above. After subtracting the $\alpha=0$ and $|\alpha|=1$ contributions from $c$, the remainder lies in $S^{m+m'-2}(U \times \mathbb{R}^n)$; the same is true for $d$. Since the $\alpha=0$ contributions cancel, the class of $c-d$ modulo $S^{m+m'-2}(U \times \mathbb{R}^n)$ is exactly the $|\alpha|=1$ difference. Hence
\begin{align*}
c-d \equiv r_{m+m'-1} \pmod{S^{m+m'-2}}.
\end{align*}
[/guided]
[/step]
[step:Replace full symbols by principal representatives]
Choose representatives $a_m \in S^m$ and $b_{m'} \in S^{m'}$ of the principal symbol classes of $A$ and $B$. Then there exist lower-order symbols
\begin{align*}
a_{\mathrm{low}}: U \times \mathbb{R}^n \to \mathbb{C}
\end{align*}
and
\begin{align*}
b_{\mathrm{low}}: U \times \mathbb{R}^n \to \mathbb{C}
\end{align*}
such that
\begin{align*}
a = a_m + a_{\mathrm{low}}, \qquad a_{\mathrm{low}} \in S^{m-1},
\end{align*}
and
\begin{align*}
b = b_{m'} + b_{\mathrm{low}}, \qquad b_{\mathrm{low}} \in S^{m'-1}.
\end{align*}
In the expression
\begin{align*}
\frac{1}{i}\sum_{j=1}^n \left(\partial_{\xi_j}a\,\partial_{x_j}b - \partial_{\xi_j}b\,\partial_{x_j}a\right),
\end{align*}
any term involving $a_{\mathrm{low}}$ has order at most $m+m'-2$, because one $\xi$-derivative falls on either $a_{\mathrm{low}} \in S^{m-1}$ or on $b \in S^{m'}$. Likewise, any term involving $b_{\mathrm{low}}$ has order at most $m+m'-2$. Hence
\begin{align*}
r_{m+m'-1} \equiv \frac{1}{i}\sum_{j=1}^n \left(\partial_{\xi_j}a_m\,\partial_{x_j}b_{m'} - \partial_{\xi_j}b_{m'}\,\partial_{x_j}a_m\right) \pmod{S^{m+m'-2}}.
\end{align*}
Since scalar multiplication is commutative, the second summand can be rewritten as
\begin{align*}
\partial_{\xi_j}b_{m'}\,\partial_{x_j}a_m = \partial_{x_j}a_m\,\partial_{\xi_j}b_{m'}.
\end{align*}
Therefore
\begin{align*}
r_{m+m'-1} \equiv \frac{1}{i}\sum_{j=1}^n \left(\partial_{\xi_j}a_m\,\partial_{x_j}b_{m'} - \partial_{x_j}a_m\,\partial_{\xi_j}b_{m'}\right) \pmod{S^{m+m'-2}}.
\end{align*}
By the stated Poisson bracket convention, this is
\begin{align*}
r_{m+m'-1} \equiv \frac{1}{i}\{a_m,b_{m'}\} \pmod{S^{m+m'-2}}.
\end{align*}
[/step]
[step:Identify the principal symbol class of the commutator]
From the preceding steps,
\begin{align*}
c-d \in S^{m+m'-1}(U \times \mathbb{R}^n)
\end{align*}
and
\begin{align*}
c-d \equiv \frac{1}{i}\{a_m,b_{m'}\} \pmod{S^{m+m'-2}}.
\end{align*}
Since $[A,B]=\operatorname{Op}(c-d)$, the class of $c-d$ in
$S^{m+m'-1}/S^{m+m'-2}$ is precisely the principal symbol class of the commutator. Thus
\begin{align*}
\sigma_{m+m'-1}([A,B]) = \frac{1}{i}\{a_m,b_{m'}\}
\end{align*}
in $S^{m+m'-1}/S^{m+m'-2}$.
If the symbols are classical, the principal symbol representatives $a_m$ and $b_{m'}$ may be chosen as the homogeneous principal parts of degrees $m$ and $m'$, respectively, so the same computation gives the displayed homogeneous principal symbol formula. This proves both the order drop and the stated principal symbol identity.
[/step]
