[proofplan]
The proof is a direct verification from the defining seminorm estimates for $S_\delta(m)$. Differentiating a symbol simply shifts the multi-indices in the defining estimate, and the fixed shifted order contributes the factor $h^{-\delta(|\alpha|+|\beta|)}$. For products, the Leibniz formula expresses each derivative of $ab$ as a finite sum of products of derivatives of $a$ and $b$; the two symbol estimates multiply, and the powers of $h$ combine to the required delta loss.
[/proofplan]
[step:Unpack the symbol estimates to be proved]
We use the standard defining estimate for $S_\delta(m)$. A function
$a:T^*\mathbb{R}^n \times (0,1] \to \mathbb{C}$
belongs to $S_\delta(m)$ precisely when, for every pair of multi-indices $\gamma,\eta \in \mathbb{N}_0^n$, there is a constant $C_{\gamma,\eta} > 0$ such that, for every $(x,\xi) \in T^*\mathbb{R}^n$ and every $h \in (0,1]$,
\begin{align*}
|\partial_x^\gamma \partial_\xi^\eta a(x,\xi;h)| \leq C_{\gamma,\eta} h^{-\delta(|\gamma|+|\eta|)} m(x,\xi).
\end{align*}
For a real number $\rho$, the notation $h^\rho S_\delta(m)$ means that $f \in h^\rho S_\delta(m)$ exactly when $h^{-\rho}f \in S_\delta(m)$. Equivalently, $f \in h^\rho S_\delta(m)$ if for every $\gamma,\eta \in \mathbb{N}_0^n$ there is a constant $C_{\gamma,\eta} > 0$ such that
\begin{align*}
|\partial_x^\gamma \partial_\xi^\eta f(x,\xi;h)| \leq C_{\gamma,\eta} h^{\rho-\delta(|\gamma|+|\eta|)} m(x,\xi).
\end{align*}
[/step]
[step:Shift the multi-indices to prove stability under differentiation]
Fix multi-indices $\alpha,\beta \in \mathbb{N}_0^n$, and define
$c_{\alpha,\beta}:T^*\mathbb{R}^n \times (0,1] \to \mathbb{C}$ by
\begin{align*}
c_{\alpha,\beta}(x,\xi;h) = \partial_x^\alpha \partial_\xi^\beta a(x,\xi;h).
\end{align*}
Let $\gamma,\eta \in \mathbb{N}_0^n$. Since $a \in S_\delta(m_1)$, the defining estimate applied with multi-indices $\alpha+\gamma$ and $\beta+\eta$ gives a constant $C_{\alpha+\gamma,\beta+\eta} > 0$ such that
\begin{align*}
|\partial_x^\gamma \partial_\xi^\eta c_{\alpha,\beta}(x,\xi;h)| = |\partial_x^{\alpha+\gamma}\partial_\xi^{\beta+\eta}a(x,\xi;h)|.
\end{align*}
Therefore
\begin{align*}
|\partial_x^\gamma \partial_\xi^\eta c_{\alpha,\beta}(x,\xi;h)| \leq C_{\alpha+\gamma,\beta+\eta} h^{-\delta(|\alpha+\gamma|+|\beta+\eta|)} m_1(x,\xi).
\end{align*}
Using $|\alpha+\gamma|=|\alpha|+|\gamma|$ and $|\beta+\eta|=|\beta|+|\eta|$, this becomes
\begin{align*}
|\partial_x^\gamma \partial_\xi^\eta c_{\alpha,\beta}(x,\xi;h)| \leq C_{\alpha+\gamma,\beta+\eta} h^{-\delta(|\alpha|+|\beta|)} h^{-\delta(|\gamma|+|\eta|)} m_1(x,\xi).
\end{align*}
This is exactly the defining estimate for
$c_{\alpha,\beta} \in h^{-\delta(|\alpha|+|\beta|)}S_\delta(m_1)$.
[/step]
[step:Apply the multi-index Leibniz formula to prove stability under multiplication]
Since $m_1$ and $m_2$ are order functions, their pointwise product $m_1m_2:T^*\mathbb{R}^n \to (0,\infty)$ is the order function defined by $(m_1m_2)(x,\xi)=m_1(x,\xi)m_2(x,\xi)$.
Let
$p:T^*\mathbb{R}^n \times (0,1] \to \mathbb{C}$
be the pointwise product defined by
\begin{align*}
p(x,\xi;h)=a(x,\xi;h)b(x,\xi;h).
\end{align*}
Fix $\alpha,\beta \in \mathbb{N}_0^n$. The multi-index Leibniz formula gives
\begin{align*}
\partial_x^\alpha \partial_\xi^\beta p(x,\xi;h)=\sum_{\mu\leq\alpha}\sum_{\nu\leq\beta}\binom{\alpha}{\mu}\binom{\beta}{\nu}(\partial_x^\mu\partial_\xi^\nu a)(x,\xi;h)(\partial_x^{\alpha-\mu}\partial_\xi^{\beta-\nu} b)(x,\xi;h).
\end{align*}
Since $a\in S_\delta(m_1)$ and $b\in S_\delta(m_2)$, for each pair $(\mu,\nu)$ appearing in the finite sum there are constants $A_{\mu,\nu}>0$ and $B_{\alpha-\mu,\beta-\nu}>0$ such that
\begin{align*}
|\partial_x^\mu\partial_\xi^\nu a(x,\xi;h)| \leq A_{\mu,\nu} h^{-\delta(|\mu|+|\nu|)}m_1(x,\xi)
\end{align*}
and
\begin{align*}
|\partial_x^{\alpha-\mu}\partial_\xi^{\beta-\nu} b(x,\xi;h)| \leq B_{\alpha-\mu,\beta-\nu} h^{-\delta(|\alpha-\mu|+|\beta-\nu|)}m_2(x,\xi).
\end{align*}
Multiplying these estimates and using
\begin{align*}
|\mu|+|\alpha-\mu|+|\nu|+|\beta-\nu|=|\alpha|+|\beta|
\end{align*}
gives
\begin{align*}
|(\partial_x^\mu\partial_\xi^\nu a)(x,\xi;h)(\partial_x^{\alpha-\mu}\partial_\xi^{\beta-\nu} b)(x,\xi;h)| \leq A_{\mu,\nu}B_{\alpha-\mu,\beta-\nu} h^{-\delta(|\alpha|+|\beta|)}m_1(x,\xi)m_2(x,\xi).
\end{align*}
Define the finite constant
\begin{align*}
C_{\alpha,\beta}=\sum_{\mu\leq\alpha}\sum_{\nu\leq\beta}\binom{\alpha}{\mu}\binom{\beta}{\nu}A_{\mu,\nu}B_{\alpha-\mu,\beta-\nu}.
\end{align*}
Taking absolute values in the Leibniz formula and applying the triangle inequality yields
\begin{align*}
|\partial_x^\alpha \partial_\xi^\beta p(x,\xi;h)| \leq C_{\alpha,\beta}h^{-\delta(|\alpha|+|\beta|)}m_1(x,\xi)m_2(x,\xi).
\end{align*}
This is the defining estimate for $p\in S_\delta(m_1m_2)$.
[guided]
We need to prove that the product has the same symbol estimates, with the product order function $m_1m_2:T^*\mathbb{R}^n \to (0,\infty)$ defined by
\begin{align*}
(m_1m_2)(x,\xi)=m_1(x,\xi)m_2(x,\xi).
\end{align*}
Define the product symbol
$p:T^*\mathbb{R}^n \times (0,1] \to \mathbb{C}$ by
\begin{align*}
p(x,\xi;h)=a(x,\xi;h)b(x,\xi;h).
\end{align*}
To show $p\in S_\delta(m_1m_2)$, we must estimate every mixed derivative $\partial_x^\alpha\partial_\xi^\beta p$. Fix multi-indices $\alpha,\beta \in \mathbb{N}_0^n$. The correct tool is the multi-index Leibniz formula, because a derivative of a product is a finite sum over all ways to distribute the $x$-derivatives and $\xi$-derivatives between the two factors:
\begin{align*}
\partial_x^\alpha \partial_\xi^\beta p(x,\xi;h)=\sum_{\mu\leq\alpha}\sum_{\nu\leq\beta}\binom{\alpha}{\mu}\binom{\beta}{\nu}(\partial_x^\mu\partial_\xi^\nu a)(x,\xi;h)(\partial_x^{\alpha-\mu}\partial_\xi^{\beta-\nu} b)(x,\xi;h).
\end{align*}
Here $\mu\leq\alpha$ means $\mu_i\leq\alpha_i$ for every component $i\in\{1,\dots,n\}$, and similarly $\nu\leq\beta$. The sums are finite because there are only finitely many such multi-indices.
Now we estimate one summand. Since $a\in S_\delta(m_1)$, the defining estimate gives a constant $A_{\mu,\nu}>0$ such that
\begin{align*}
|\partial_x^\mu\partial_\xi^\nu a(x,\xi;h)| \leq A_{\mu,\nu} h^{-\delta(|\mu|+|\nu|)}m_1(x,\xi).
\end{align*}
Since $b\in S_\delta(m_2)$, the defining estimate gives a constant $B_{\alpha-\mu,\beta-\nu}>0$ such that
\begin{align*}
|\partial_x^{\alpha-\mu}\partial_\xi^{\beta-\nu} b(x,\xi;h)| \leq B_{\alpha-\mu,\beta-\nu} h^{-\delta(|\alpha-\mu|+|\beta-\nu|)}m_2(x,\xi).
\end{align*}
Multiplying the two inequalities gives
\begin{align*}
|(\partial_x^\mu\partial_\xi^\nu a)(x,\xi;h)(\partial_x^{\alpha-\mu}\partial_\xi^{\beta-\nu} b)(x,\xi;h)| \leq A_{\mu,\nu}B_{\alpha-\mu,\beta-\nu} h^{-\delta(|\mu|+|\nu|+|\alpha-\mu|+|\beta-\nu|)}m_1(x,\xi)m_2(x,\xi).
\end{align*}
The exponent is exactly the exponent required for an $(\alpha,\beta)$ derivative, because multi-index length is additive under this splitting:
\begin{align*}
|\mu|+|\alpha-\mu|+|\nu|+|\beta-\nu|=|\alpha|+|\beta|.
\end{align*}
Thus each summand is bounded by a constant times
\begin{align*}
h^{-\delta(|\alpha|+|\beta|)}m_1(x,\xi)m_2(x,\xi).
\end{align*}
Because the Leibniz formula contains only finitely many summands, we may collect their constants into one finite constant:
\begin{align*}
C_{\alpha,\beta}=\sum_{\mu\leq\alpha}\sum_{\nu\leq\beta}\binom{\alpha}{\mu}\binom{\beta}{\nu}A_{\mu,\nu}B_{\alpha-\mu,\beta-\nu}.
\end{align*}
Taking absolute values in the Leibniz formula and applying the triangle inequality gives
\begin{align*}
|\partial_x^\alpha \partial_\xi^\beta p(x,\xi;h)| \leq C_{\alpha,\beta}h^{-\delta(|\alpha|+|\beta|)}m_1(x,\xi)m_2(x,\xi).
\end{align*}
This is precisely the defining estimate for $p\in S_\delta(m_1m_2)$. Since $\alpha$ and $\beta$ were arbitrary, the product $ab$ belongs to $S_\delta(m_1m_2)$.
[/guided]
[/step]
[step:Conclude both stability assertions]
The differentiation estimate was proved for arbitrary $\alpha,\beta \in \mathbb{N}_0^n$, so every fixed mixed derivative $\partial_x^\alpha\partial_\xi^\beta a$ lies in $h^{-\delta(|\alpha|+|\beta|)}S_\delta(m_1)$. The product estimate was proved for arbitrary derivatives of $ab$, so $ab\in S_\delta(m_1m_2)$. These are exactly the two asserted stability properties.
[/step]
