= Solution
For $a\in G(A)$, any $b\in A$ with $\|a^{-1}\|\|b-a\|<1$ has the factorization
$$
b=a\bigl(1+a^{-1}(b-a)\bigr).
$$
The second factor has a <Neumann series> inverse in $A$, so $b\in G(A)$. Hence the <group of invertible elements of a Banach algebra> $G(A)$ is open in $A$.
For $a\in G_B(A)$ use its inverse in $B$ instead. The same small-perturbation argument proves invertibility of $b$ in $B$ whenever $b\in A$ and $\|a^{-1}\|_B\|b-a\|_B<1$. Therefore \b[both $G(A)$ and $G_B(A)$ are open subsets of $A$]; in the second argument the inverse need not lie in $A$.
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