[proofplan] The strict topology on $M(A)$ is defined by testing multiplication on the left and on the right against each fixed element of $A$. Thus it is enough to fix $a\in A$ and prove that the two norms $\|(e_i-1_{M(A)})a\|_A$ and $\|a(e_i-1_{M(A)})\|_A$ tend to zero. These two expressions are exactly the two-sided approximate identity relations for $(e_i)_{i\in I}$. [/proofplan]

[step:Recall the strict seminorms determined by elements of $A$] For each $a\in A$, define seminorms \begin{align*} p_a:M(A)\to [0,\infty),\qquad p_a(m)=\|ma\|_A \end{align*} and \begin{align*} q_a:M(A)\to [0,\infty),\qquad q_a(m)=\|am\|_A. \end{align*} The strict topology on $M(A)$ is the locally convex topology generated by the family of seminorms $\{p_a,q_a:a\in A\}$. Therefore a net $(m_i)_{i\in I}$ in $M(A)$ converges strictly to $m\in M(A)$ if and only if, for every $a\in A$, \begin{align*} \|m_i a-ma\|_A\to 0 \end{align*} and \begin{align*} \|a m_i-a m\|_A\to 0. \end{align*} [/step]

[step:Use the approximate identity property on both sides]Fix $a\in A$. Since $(e_i)_{i\in I}$ is a two-sided approximate identity for $A$, the defining property gives \begin{align*} \|e_i a-a\|_A\to 0 \end{align*} and \begin{align*} \|a e_i-a\|_A\to 0. \end{align*}[/step]

[guided]Fix an element $a\in A$. The strict topology asks us to test convergence after multiplying by this fixed $a$ on both sides, so this is the only element that needs to be held fixed in the argument. Because $(e_i)_{i\in I}$ is a two-sided approximate identity for $A$, it satisfies two norm convergence conditions for every element of $A$: left multiplication by $e_i$ converges to the identity operation on $A$, and right multiplication by $e_i$ converges to the identity operation on $A$. Applied to the present element $a$, these conditions are precisely \begin{align*} \|e_i a-a\|_A\to 0 \end{align*} and \begin{align*} \|a e_i-a\|_A\to 0. \end{align*} The contractive hypothesis $\|e_i\|_A\le 1$ is part of the standard form of the approximate identity, but this particular convergence argument uses only the two-sided approximation property.[/guided]

[step:Identify these convergences with strict convergence to $1_{M(A)}$] Under the canonical embedding $A\subset M(A)$, each $e_i$ is an element of $M(A)$, and $1_{M(A)}$ is the identity multiplier. Hence, for the fixed $a\in A$, \begin{align*} (e_i-1_{M(A)})a=e_i a-a \end{align*} and \begin{align*} a(e_i-1_{M(A)})=a e_i-a. \end{align*} Taking $A$-norms and using the two limits from the previous step gives \begin{align*} p_a(e_i-1_{M(A)})=\|(e_i-1_{M(A)})a\|_A=\|e_i a-a\|_A\to 0 \end{align*} and \begin{align*} q_a(e_i-1_{M(A)})=\|a(e_i-1_{M(A)})\|_A=\|a e_i-a\|_A\to 0. \end{align*} Since $a\in A$ was arbitrary, every strict seminorm of $e_i-1_{M(A)}$ tends to zero. Therefore $(e_i)_{i\in I}$ converges strictly to $1_{M(A)}$ in $M(A)$. [/step]