[proofplan]
The proof uses only the definition of residual membership and the elementary multiplicative stability of symbol estimates. For each prescribed power $h^N$, write the residual symbol $r$ as $h^N r_N$ with $r_N$ still in the same symbol class. Leibniz' rule shows that multiplying $r_N$ by $a$ produces a symbol controlled by the product order function $m_1m_2$, so $ar$ has the same arbitrary $h^N$ gain. Since $N$ is arbitrary, $ar$ is residual in the product class.
[/proofplan]
[step:Convert residual decay of $r$ into an arbitrary fixed power of $h$]
Fix $N \in \mathbb{N}_0$. Since $r \in h^\infty S(m_1)$, by definition there exists a symbol
\begin{align*}
r_N:T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C}
\end{align*}
with $r_N \in S(m_1)$ such that
\begin{align*}
r = h^N r_N.
\end{align*}
Therefore
\begin{align*}
ar = h^N a r_N.
\end{align*}
It remains to verify that $a r_N \in S(m_1m_2)$.
[guided]
The phrase $r \in h^\infty S(m_1)$ means that $r$ gains every power of the semiclassical parameter $h$ while retaining the $S(m_1)$ symbol estimates after removing that power. Thus, once we choose a specific integer $N \in \mathbb{N}_0$, the definition gives a symbol
\begin{align*}
r_N:T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C}
\end{align*}
such that $r_N \in S(m_1)$ and
\begin{align*}
r = h^N r_N.
\end{align*}
Multiplying by $a$ gives
\begin{align*}
ar = a h^N r_N = h^N a r_N.
\end{align*}
So the whole problem has been reduced to one ordinary symbol-class question: if $a \in S(m_2)$ and $r_N \in S(m_1)$, then their product should lie in $S(m_1m_2)$. Once that is proved, the displayed identity shows that $ar$ lies in $h^N S(m_1m_2)$ for the chosen $N$.
[/guided]
[/step]
[step:Use Leibniz' rule to place the product in $S(m_1m_2)$]
Let $\alpha,\beta \in \mathbb{N}_0^n$ be multiindices. Since $a \in S(m_2)$ and $r_N \in S(m_1)$, for every pair of multiindices $\gamma,\eta \in \mathbb{N}_0^n$ there exist constants $A_{\gamma,\eta}>0$ and $R_{\gamma,\eta}>0$ such that, for all $(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]$,
\begin{align*}
|\partial_x^\gamma \partial_\xi^\eta a(x,\xi;h)| \leq A_{\gamma,\eta} m_2(x,\xi)
\end{align*}
and
\begin{align*}
|\partial_x^\gamma \partial_\xi^\eta r_N(x,\xi;h)| \leq R_{\gamma,\eta} m_1(x,\xi).
\end{align*}
By Leibniz' rule,
\begin{align*}
\partial_x^\alpha \partial_\xi^\beta(a r_N) = \sum_{\gamma \leq \alpha} \sum_{\eta \leq \beta} \binom{\alpha}{\gamma} \binom{\beta}{\eta} (\partial_x^\gamma \partial_\xi^\eta a)(\partial_x^{\alpha-\gamma} \partial_\xi^{\beta-\eta} r_N)
\end{align*}
Define the finite Leibniz constant
\begin{align*}
C_{\alpha,\beta,N} = \sum_{\gamma \leq \alpha} \sum_{\eta \leq \beta} \binom{\alpha}{\gamma} \binom{\beta}{\eta} A_{\gamma,\eta} R_{\alpha-\gamma,\beta-\eta}.
\end{align*}
Then, for all $(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]$,
\begin{align*}
|\partial_x^\alpha \partial_\xi^\beta(a r_N)(x,\xi;h)| \leq C_{\alpha,\beta,N} m_1(x,\xi)m_2(x,\xi).
\end{align*}
Thus $a r_N \in S(m_1m_2)$.
[guided]
We now prove the ordinary product estimate that was isolated in the previous step. Fix multiindices $\alpha,\beta \in \mathbb{N}_0^n$. The definition of $a \in S(m_2)$ gives constants $A_{\gamma,\eta}>0$ for every $\gamma,\eta \in \mathbb{N}_0^n$ such that
\begin{align*}
|\partial_x^\gamma \partial_\xi^\eta a(x,\xi;h)| \leq A_{\gamma,\eta} m_2(x,\xi)
\end{align*}
for all $(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]$. Likewise, since $r_N \in S(m_1)$, there are constants $R_{\gamma,\eta}>0$ such that
\begin{align*}
|\partial_x^\gamma \partial_\xi^\eta r_N(x,\xi;h)| \leq R_{\gamma,\eta} m_1(x,\xi).
\end{align*}
Leibniz' rule distributes the $\alpha$ derivatives in $x$ and the $\beta$ derivatives in $\xi$ between the two factors:
\begin{align*}
\partial_x^\alpha \partial_\xi^\beta(a r_N) = \sum_{\gamma \leq \alpha} \sum_{\eta \leq \beta} \binom{\alpha}{\gamma} \binom{\beta}{\eta} (\partial_x^\gamma \partial_\xi^\eta a)(\partial_x^{\alpha-\gamma} \partial_\xi^{\beta-\eta} r_N).
\end{align*}
Define the finite Leibniz constant by
\begin{align*}
C_{\alpha,\beta,N} = \sum_{\gamma \leq \alpha} \sum_{\eta \leq \beta} \binom{\alpha}{\gamma} \binom{\beta}{\eta} A_{\gamma,\eta} R_{\alpha-\gamma,\beta-\eta}.
\end{align*}
Taking absolute values and applying the two symbol estimates to each summand gives, for all $(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]$,
\begin{align*}
|\partial_x^\alpha \partial_\xi^\beta(a r_N)(x,\xi;h)| \leq C_{\alpha,\beta,N} m_1(x,\xi)m_2(x,\xi).
\end{align*}
Because this estimate holds for every pair $\alpha,\beta$, it is exactly the defining estimate for $a r_N \in S(m_1m_2)$.
[/guided]
[/step]
[step:Conclude residual membership in the product class]
For the fixed $N \in \mathbb{N}_0$, we have shown that
\begin{align*}
ar = h^N c_N
\end{align*}
with
\begin{align*}
c_N := a r_N \in S(m_1m_2).
\end{align*}
Therefore $ar \in h^N S(m_1m_2)$. Since $N \in \mathbb{N}_0$ was arbitrary,
\begin{align*}
ar \in \bigcap_{N=0}^{\infty} h^N S(m_1m_2) = h^\infty S(m_1m_2).
\end{align*}
[/step]
[step:Repeat the argument with the $S_\delta$ symbol estimates]
Fix $\delta$ satisfying
\begin{align*}
0 \leq \delta < \frac{1}{2},
\end{align*}
and assume $r \in h^\infty S_\delta(m_1)$ and $a \in S_\delta(m_2)$. For each $N \in \mathbb{N}_0$, choose
\begin{align*}
r_N:T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C}
\end{align*}
with $r_N \in S_\delta(m_1)$ and $r = h^N r_N$. The $S_\delta$ estimates give constants $A_{\gamma,\eta}>0$ and $R_{\gamma,\eta}>0$ such that
\begin{align*}
|\partial_x^\gamma \partial_\xi^\eta a(x,\xi;h)| \leq A_{\gamma,\eta} h^{-\delta(|\gamma|+|\eta|)} m_2(x,\xi)
\end{align*}
and
\begin{align*}
|\partial_x^\gamma \partial_\xi^\eta r_N(x,\xi;h)| \leq R_{\gamma,\eta} h^{-\delta(|\gamma|+|\eta|)} m_1(x,\xi).
\end{align*}
Applying Leibniz' rule to $\partial_x^\alpha \partial_\xi^\beta(a r_N)$, each summand has total derivative order $|\alpha|+|\beta|$, so the powers of $h$ combine to give exactly the factor
\begin{align*}
h^{-\delta(|\alpha|+|\beta|)}.
\end{align*}
Define
\begin{align*}
C_{\alpha,\beta,N,\delta} = \sum_{\gamma \leq \alpha} \sum_{\eta \leq \beta} \binom{\alpha}{\gamma}\binom{\beta}{\eta} A_{\gamma,\eta}R_{\alpha-\gamma,\beta-\eta}.
\end{align*}
Therefore, for all $(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]$,
\begin{align*}
|\partial_x^\alpha \partial_\xi^\beta(a r_N)(x,\xi;h)| \leq C_{\alpha,\beta,N,\delta} h^{-\delta(|\alpha|+|\beta|)} m_1(x,\xi)m_2(x,\xi).
\end{align*}
Hence $a r_N \in S_\delta(m_1m_2)$, and so
\begin{align*}
ar \in h^N S_\delta(m_1m_2).
\end{align*}
Since $N$ was arbitrary,
\begin{align*}
ar \in h^\infty S_\delta(m_1m_2).
\end{align*}
This proves both assertions.
[guided]
The $S_\delta$ case follows the same structure, but we must track the derivative loss in powers of $h$. Fix $\delta$ with
\begin{align*}
0 \leq \delta < \frac{1}{2},
\end{align*}
and assume $r \in h^\infty S_\delta(m_1)$ and $a \in S_\delta(m_2)$. For a chosen $N \in \mathbb{N}_0$, residual membership gives
\begin{align*}
r_N:T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C}
\end{align*}
with $r_N \in S_\delta(m_1)$ and $r=h^N r_N$.
For each pair of multiindices $\gamma,\eta \in \mathbb{N}_0^n$, the $S_\delta$ estimates provide constants $A_{\gamma,\eta}>0$ and $R_{\gamma,\eta}>0$ such that
\begin{align*}
|\partial_x^\gamma \partial_\xi^\eta a(x,\xi;h)| \leq A_{\gamma,\eta} h^{-\delta(|\gamma|+|\eta|)}m_2(x,\xi)
\end{align*}
and
\begin{align*}
|\partial_x^\gamma \partial_\xi^\eta r_N(x,\xi;h)| \leq R_{\gamma,\eta} h^{-\delta(|\gamma|+|\eta|)}m_1(x,\xi).
\end{align*}
Apply Leibniz' rule to $\partial_x^\alpha\partial_\xi^\beta(a r_N)$. In the summand indexed by $\gamma \leq \alpha$ and $\eta \leq \beta$, the two derivative orders add as
\begin{align*}
|\gamma|+|\eta|+|\alpha-\gamma|+|\beta-\eta|=|\alpha|+|\beta|.
\end{align*}
Thus the two $h$-losses multiply to exactly $h^{-\delta(|\alpha|+|\beta|)}$, not to a worse power. Define the finite Leibniz constant
\begin{align*}
C_{\alpha,\beta,N,\delta} = \sum_{\gamma \leq \alpha} \sum_{\eta \leq \beta} \binom{\alpha}{\gamma}\binom{\beta}{\eta} A_{\gamma,\eta}R_{\alpha-\gamma,\beta-\eta}.
\end{align*}
Therefore, for all $(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]$,
\begin{align*}
|\partial_x^\alpha \partial_\xi^\beta(a r_N)(x,\xi;h)| \leq C_{\alpha,\beta,N,\delta} h^{-\delta(|\alpha|+|\beta|)}m_1(x,\xi)m_2(x,\xi).
\end{align*}
Hence $a r_N \in S_\delta(m_1m_2)$, so $ar=h^N a r_N \in h^N S_\delta(m_1m_2)$. Since $N$ was arbitrary, $ar \in h^\infty S_\delta(m_1m_2)$.
[/guided]
[/step]
