Remark 6.1.8.label One may arrive at the Banach algebra structure of $A_{I}$ by requiring
\begin{align*}(x,0)(y,0)=(xy,0)&&(0,\lam)(0,\mu)=(0,\lam\mu)&&(x,0)(0,\mu)=(\mu x,0)&&(0,\lam)(y,0)=(\lam y,0)\end{align*}
Equivalently we may require $(0,1)\in A\op \com$ to be the identity and multiplication is bilinear since
\begin{align*}(0,\lam)(y,0)=\lam(0,1)(y,0)=\lam(y,0)=(\lam y,0)\end{align*}
Lastly we remark that
\begin{align*}(0,\mu)=(0,\mu)+(0,0)\implies (0,\mu)(x,\lam)=(\mu x,\mu\lam)\end{align*}