I think the OP needs to give a precise formulation of what he/she means by "a Banach space inside a Banach algebra", because this affects the answers to the questions in non-trivial ways.
Contrary to what the OP seems to claim in comments to Akhil Matthew's answer, if $B$ is not closed in $A$, then it can be an inverse-closed subspace without the given Banach norm on $B$ being submultiplicative. However, by Akhil's argument, $B$ is a subalgebra of $A$.
It is still not clear to me what precisely is meant by (2). So I shall just point out for the record that if we let $A({\mathbb T})$ be the space of all continuous functions ${\mathbb T}\to {\mathbb C}$ with absolutely convergent Fourier series, equipped with pointwise product, this is a Banach algebra and it is a unital subalgebra of $C({\mathbb T})$. Clearly the norm on $A({\mathbb T})$ is not equivalent to the sup-norm inherited from $C({\mathbb T})$; but $A({\mathbb T})$ is inverse-closed in $C({\mathbb T})$, by Gelfand's version of Wiener's $1/f$ lemma.
Note also in (3) that one can have commutative, semisimple, unital Banach algebras $A$ and $B$, and an injective, contractive, unital algebra homomorphism $B\to A$ with dense range, such that $A$ and $B$ have different maximal ideal spaces. The example I have in mind is due, I think, to Honary, but I am out of the office right now and can't look this up.