= Solution
In a unital <Banach algebra>, the <Neumann series> gives $(1-u)^{-1}=\sum_{k=0}^\infty u^k$ whenever $\|u\|<1$: the series converges in <norm> by <submultiplicativity> and <completeness>, and multiplication of its partial sums on either side gives $1-u^{N+1}\to1$. Consequently, if $g$ is invertible and $\|g^{-1}\|\|h-g\|<1$, write
$$
h=g(1+v),\qquad v=g^{-1}(h-g).
$$
The second factor has a <Neumann series> inverse, so $h$ is invertible. This exhibits a ball around each $g$ in the <group of invertible elements of a Banach algebra>, proving that group is an <open set>.
The same series proves <continuity of inversion in a Banach algebra>, with the useful quantitative estimate
$$
\|h^{-1}-g^{-1}\|
=\|[(1+v)^{-1}-1]g^{-1}\|
\le\frac{\|g^{-1}\|^2\|h-g\|}{1-\|g^{-1}\|\|h-g\|}.
$$
The right side tends to zero as $h\to g$. Since applying inversion twice gives the original element, it is a <bijection> whose inverse is the same <continuous> map. Thus \b[inversion is a homeomorphism of the open group of invertibles onto itself].
For the final unheaded request, define
$$
E=\{\lambda\in\mathbb C:1-\lambda a\text{ is invertible}\}.
$$
It is open by the preceding argument and contains $0$. If $\lambda_0$ were a <boundary> point, a sequence $\lambda_n\in E$ would converge to $\lambda_0$, while openness gives $\lambda_0\notin E$. Then $1-\lambda_na\to1-\lambda_0a$ is a noninvertible limit of invertibles. The one-sided-inverse argument in part (ii) shows that this limit has neither a <left inverse of an algebra element> nor a <right inverse of an algebra element>, contradicting the hypothesis. Therefore $E$ has empty <boundary>; it is both open and <closed> in the <connected> <complex plane>, and $E=\mathbb C$.
For every nonzero $\mu$, put $\lambda=\mu^{-1}$. The identity $\mu1-a=\mu(1-\lambda a)$ proves invertibility, so the <spectrum of an element> $a$ is contained in $\{0\}$. It is nonempty: if the <resolvent of an element> existed for all $\mu\in\mathbb C$, it would be an entire algebra-valued function tending to zero at infinity by the <Neumann series>. Applying any <bounded linear functional> and then the <Liouville theorem> would make it zero; the <Hahn-Banach theorem> separates points and would force the resolvent itself to be zero, contradicting its inverse equation. Hence
$$
\boxed{\sigma_A(a)=\{0\}.}
$$
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