 Solution

For $a\in G(A)$, any $b\in A$ with $\Vert a^{-1}\Vert \Vert b-a\Vert <1$ has the factorizationThe 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 $\Vert a^{-1}{\Vert }_B\Vert b-a{\Vert }_B<1$. Therefore both $G(A)$ and $G_B(A)$ are open subsets of $A$; in the second argument the inverse need not lie in $A$.