If A is a $C^*$ algebra, then there is a $C^*$ algebra $A_1$ with an identity such that $A_1$ contains A as an ideal. This is in the book Functional analysis by John B. Conway. Chapter 9 prop: 1.9.
Basically in the book, he takes $A_1=\{a+\alpha \}$ where $(a+\alpha )$ is just a formal sum. And defined multiplication and addition as:
i) $(a+\alpha )(b+\beta )=(ab+\alpha b+\beta a+\alpha \beta )$
ii) $(a+\alpha )+(b+\beta )=(a+b)+(\alpha +\beta )$
And norm as:$$\|a+\alpha \|=\sup \{ \|ax+\alpha x\|:x\in A,\|x\|\leq 1\}.$$ I am unable to prove that $A_1$ is a Banach algebra. I am trying to show if $\|a+\alpha \|=\|a\|+|\alpha |$ or not but unable to do so.
 A: First of all, the norm of $a+\alpha$ is not $\|a\|+|\alpha|$ in general. That's the norm one uses to make the unitization a Banach algebra, but it is not a C$^*$-norm.
In principle you have that $A_1$ is a $*$-algebra, and you can show that the norm satisfies the C$^*$-identity. Then you can consider its completion $\overline{A_1}$, which ends up being $A_1$ but we still don't know that. Using Hahn-Banach, there exists a bounded linear functional $\varphi$ on $A_1$ such that $\varphi|_A=0$, $\varphi(1)=1$. Now suppose that you have a Cauchy sequence $\{a_n+\alpha_n\}$. We have
$$
|\alpha_n-\alpha_m|=|\varphi(a_n+\alpha_n-(a_m+\alpha_m))|
\leq\|\varphi\|\,\|a_n+\alpha_n-(a_m+\alpha_m)\|,
$$
and it follows that $\{\alpha_n\}$ is Cauchy. From the triangle inequality,
$$
\|a_n-a_m\|\leq\|a_n+\alpha_n-(a_m+\alpha_m)\|+\|\alpha_n-\alpha_m\|
$$
and, since both sequences on the right-hand-side are Cauchy, we get that $\{a_n\}$ is Cauchy.
So $a_n\to a $ in $A$, $\alpha_n\to\alpha$ in $\mathbb C$, and
$$
\|a_n+\alpha-(a+\alpha)\|\leq\|a_n-a\|+|\alpha_n-\alpha|\to0. 
$$
