Solution
= Solution
Let $c=(1-ab)^{-1}\in A$. From the two inverse identities we have $cab=c-1=abc$. The candidate inverse is
$$
\boxed{(1-ba)^{-1}=1+bca.}
$$
Indeed
$$
(1-ba)(1+bca)=1+b(c-1-abc)a=1,
$$
and
$$
(1+bca)(1-ba)=1+b(c-1-cab)a=1.
$$
Both products have been checked, so this is valid in a noncommutative <Banach algebra>. It is the identity underlying the <nonzero spectra of products in opposite orders> result.