When reading Murphy's book $C^*$ algebras and operator theory, I found the Lemma 1.3.2 a bit strange. The original lemmas is as the following:
If $I$ is a modular maximal ideal of a unital abelian Banach algebra $A$, then $A/I$ is a field.
I want to ask that is the assumption " Abelian " redundant? Since by Gelfand-Mazur theorem, we know that if all nonzero elements in $A$ are invertible then $A = \mathbb{C}\cdot1$. And here, we have a modular maximal ideal $I$ which is closed, then the quotient $A/I$ is still complete hence still a Banach algebra, also, since $I$ is maximal, all nonzero elements in $A/I$ are invertible then apply the Gelfand-Mazur theorem, we can conclude that $A/I \cong \mathbb{C}$. I cannot see where the assumption "abelian" is applied.
In the proof of I.3.2, you need "abelian" to have that $J=\pi(a)\,(A/I)$ is an ideal and not just a right-ideal.
In the non-abelian case, the assertion is very non-true. For instance consider $B(H)$ with $H$ infinite-dimensional. Then the only ideal is $K(H)$ and $B(H)/K(H)$ is the (non-separable, very non-abelian) Calkin algebra. And there is no theory needed to understand this: take $P$ an infinite projection such that $I-P$ is also infinite. Then $P$ maps to a non-invertible element in the Calkin algebra, since $P(I-P)=0$.
There are lots of non-abelian simple C$^*$-algebras. For instance $M_n(\mathbb C)$ for any $n$, the Cuntz algebra $\mathscr O_n$ for any $n$, the reduced free group algebra $C^*(\mathbb F_n)$ for any $n$, etc. In all of those the only ideal is $\{0\}$, so $A/I=A$.