The following is taken from Dales' Banach Algebras and Automatic Continuity;
For reference, $A^\bullet=\{a\in A:a\neq 0\}$, $A^2=\{a^2:a\in A\}$, $AA=\{ab:a,b\in A\}$, $aA=\{ab:b\in A\}$, etc.
None of the algebras are assumed to be unital, unless stated otherwise.
I'm struggling to follow the proof of 1.3.52(i):
It's claimed that $A^2\neq A$ implies $I\neq A$. Are we using the definition of simplicity here? Should it perhaps be $A^2\neq 0$? Or should the definition of simplicity be amended to $A^2\neq A$?
Why does $AaA\neq 0$ imply $AaA=A$? If this is a direct appeal to simplicity, I don't see why $AaA$ is an ideal.
Are you sure about your notation? On p. 29 of the book, for a subset $S$ of $A$, the author defines $S^{[2]}$ to be the set of all products $xy$ with $x,y\in S$ and defines $S^2$ to be the linear span of $S^{[2]}$. More generally, the author defines $ST$ to be the linear span of all products $st$, $s\in S,t\in T$. This follows standard notation in ring theory. In particular, $A^2$ is always an ideal. For a ring with no ideals save $0$ and $A$, the conditions $A^2=A$ and $A^2\ne 0$ are equivalent.
If $a\in A$, the (usual and also this book's) meaning of $AaA$ is the ideal generated by all products $xay$, $x,y\in A$, that is, the set of elements $\sum_{j=1}^n x_jay_j$. This set is an ideal, so if $AaA\ne0$, it must be $A$ by simplicity. Moreover, $I=\{\,a\in A\mid AaA=0\,\}$ is an ideal, so it is either $0$ or $A$. If it is $A$, then $A^3=0$, which contradicts the assumption that $A^2=A$. Thus $I=0$.