It is a fact that the character space of a commutative $C^\ast$-algebra is non-empty. Consider the following proof, also included here:
Proof:
My question is: could someone please explain to me why the latter part involving the unitisation and the restrtiction of the character to $A$ is necessary?
Concretely, my own proof would be the following and I don't know why it might not be correct:
Recall that in a commutative non-unital Banach algebra $A$ we have $\sigma (a) = \{ \tau (a) \mid \tau \in \Omega (A) \} \cup \{0\}$. Since there exists $\tau $ such that $|\tau (a) | = \|a\| > 0$ it follows immediately that $\tau \in \Omega (A)$ is a non-zero character which proves the claim.
The unitisation is necessary to show the existence of characters: you need the existence of maximal ideals, which you can only assure they exist in unital abelian algebras.
I don't know where the problem is; I have to admit I know nothing about Banach algebra unitisations. Consider this example: let $A=\ell^1(\mathbb N)$, with the pointwise addition and multiplication. It is not hard to see that the characters are precisely the maps $$ a\longmapsto a(n) $$ for a fixed $n$. But then if for example you consider the sequence $(2^{-n})$, then $|\tau(a)|\leq1/2$ for all characters, while $\|a\|_1=1$. So it looks like there is no character with $|\tau(a)|=\|a\|$.