Let $A$ be a unital $C^*$-algebra.
Question: Is $A$ isomorphic to the unitization of a $C^*$-algebra $A_0$?
Clarification: Here, I am using the functorial definition of unitization. The unitization of $A_0$ is the algebra with underlying vector space $A_0\oplus\mathbb C$ and norm
$$||(a,\lambda)\|=\max\Big\{|\lambda|,\sup_{\|b\|=1}\{\|ab+\lambda b\|\}\Big\}.$$
Thoughts: Suppose $A$ is commutative. Then by Gelfand duality, $A$ is isomorphic to $C(X)$ for some compact Hausdorff space $X$. So in the commutative case it would suffice that any such $X$ is the one-point compactification of some space $X_0$; but I'm not even sure if this is true.
Best Answer
No. The algebra $A_0$ in question would necessarily be an ideal -- the multiplication you have on the unitization is $$(a,\mu)(b,\nu) = (ab + \mu b + \nu a, \mu \nu).$$ Thus if $\mu$ or $\nu$ is 0, then the right component will be 0, i.e., $A_0$ is an ideal. Are there C*-algebras without proper ideals? (try $M_n$)
As for the commutative case, see Which spaces can be seen as to be the one-point compactification of some other space.
Of course, the unitization you describe is a "minimal" unitization. However every unital C*-algebra $A$ is in fact the multiplier algebra of itself (which is in some sense a "maximal unitization" -- it corresponds to the Stone–Čech compactification in the commutative case).