Lemma 6.1.7.label Let $A$ be a non-unital $*$-algebra then there exists an embedding of $A$ into a unital $*$-algebra $A_{I}$ as an ideal.
Proof. Let $A_{I}\define A\op \com$ as a linear space with the Banach algebra structure
\begin{gather*}(x,\lam)(y,\mu)=(xy+\mu x+\lam y,\lam\mu)\\ (x,\lam)^{*}=(x^{*},\cl{\lam})\\ \norm{(x,\lam)}{}=\norm{x}{}+\abs{\lam}\end{gather*}
for every $(x,\lam),(y,\mu)\in A\op \com$. The map $x\in A\mto (x,0)\in A_{I}$ is an isometric isomorphism and the element $(0,1)$ is the identity of $A_{I}$. Identifying each $x\in A$ and $(x,0)\in A_{I}$, we write $(x,\lam)=x+\lam 1\in A_{I}$. Under the identification, $A$ is an ideal of $A_{I}$.$\square$