Let $A$ be a unital Banach Algebra. Let $B$ be a maximal Abelian Banach sub-algebra containing 1. I want to show that $Inv(B)=B\cap Inv(A)$, where $Inv(P)$ is the set of invertible elements in $P$.
One inclusion is trivial. For the other inclusion let $x\in B\cap Inv(A)$. We know that $b^{-1}$ exists in $A$, we need to show that it lies in $B$. I believe the idea should be to try and use $b^{-1}$ to generate an abelian sub-algebra bigger than $B$. This contradicts the maximality of $B$ and we are done.
I do not see how to generate such an sub algebra however, it seems to me that $B+b^{-1}\mathbb C$ is not a sub algebra of $A$.
Let $b\in B$; then for any $a\in B$, $ab=ba$, so $$b^{-1}a=ab^{-1}$$ This implies that the set of polynomials generated by $B$ and $b^{-1}$ is an abelian subalgebra of $A$; each element takes the form $$\sum_{i=0}^na_ib^{-i}$$ By the above identity, the product of such elements remains in the subalgebra and commutes. Since this algebra contains $B$, but $B$ is maximal, it follows that $b^{-1}\in B$.