Which Banach algebras are group algebras?

A characterization of the Banach algebras which are isometrically *-isomorphic to $L^1(G)$ for some $G$ was given by P. L. Patterson, Characterization of algebras arizing from locally compact groups, Trans. Amer. Math. Soc. 329 (1992), 489-506. This paper contains references to earlier ones dealing with special classes of groups, in particular Abelian and compact ones.

[This is too long to be a comment, but is at best a partial answer; I just thought I'd write it down while I have spare time.]

Regarding the 2nd part of Question 1: you need to impose some extra hypotheses to capture the "real" question I suspect you are interested in. Otherwise, just take $C[0,1]$, which is certainly abelian, symmetric as a *-algebra, and semisimple. This cannot be isomorphic to $L^1$ of anything, even as a Banach space. (To give a bit more detail: every map from $C[0,1]$ to $L^1$ is weakly compact, so if the two were isomorphic as Banach spaces then the identity map on $C[0,1]$ would be weakly compact, implying $C[0,1]$ is reflexive which is not the case.) If you want non C*-examples then $C^1[0,1]$ will also do, for much the same reasons.

If you're after a counter-example where the underlying Banach space looks like $L^1$ or $\ell^1$, then I suggest the convolution algebra $\ell^1(S)$ where $S$ is an infinite semilattice (=commutative semigroup in which all elements are idempotent); the involution is just complex conjugation. Semisimplicity is proved in an old paper of Hewitt and Zuckerman (I am not sure of the spelling of the 2nd author's name, and as I am out of the office can't check right now). Off the top of my head, the quickest, if not the most direct, way to show that such an algebra is not isomorphic to an abelian group algebra, is to use the fact (Duncan & Namioka, late 1970s) that $\ell^1(S)$ for such $S$ is never amenable, while every abelian $L^1$-group algebra is amenable (Johnson, 1972). There ought to be a simper way to distinguish the two classes of algebra, though.

Update. If you are interested only in isometric isomorphism of Banach algebras then there is a much easier way to show $\ell^1(S)$ is not a group algebra. For if $\theta: \ell^1(S)\to L^1(G)$ is an isometric algebra isomorphism, then by examining where extreme points of the unit ball go, you find that $G$ must be discrete and for each $x\in S$, $\theta(\delta_x)=\delta_{\phi(x)}$ where $\phi:S \to G$ is an isomorphism of semigroups. Now $S$ has lots of idempotents, $G$ has only one, and we have reduced to absurdity as they say.