Suppose $A$ is a commutative Banach algebra with identity and $I\subset A$ is an ideal. If there is a unique maximal ideal $M$ with $I\subset M$ does it follow that $I=M$? Or that $I$ is dense in $M$?
An answer assuming $A$ is semi-simple would be enough; in fact the case I'm really interested in is $A=L^1(G)$ where $G$ is a discrete abelian group. (So this seems related to Wiener's Tauberian Theorem; if I'm recalling the definition correctly WTT says that $\emptyset\subset\hat G$ is a set of spectral synthesis, while the current question asks whether singletons are sets of spectral synthesis.)
As for your first question the answer is no, as there exist unital, commutative Banach algebras that are also local rings. For example, consider the unitisation of
$$R = \{f\in L_0((0,\infty))\colon \int\limits_0^\infty |f(t)|e^{-t^2}\, {\rm d}t <\infty\},$$ endowed with the convolution product:
$$(f\ast g)(t) = \int\limits_0^t f(t-s)g(s)\,{\rm d}s.$$
So $\{0\}$ is contained in $R$, the unique maximal ideal of $A$.
I will reply for the second part later.