1. A Banach space with all products taken to be zero. Then every element is quasi invertible.

2.The Banach space L¹([0,1]) with the product (fg)(x) = ò₀^x f(x-y)g(y)dy has f_°(t) = t , 0 £ t £ 1 as a generator since f_°ⁿ(t) = [( t^n-1)/( (n-1)!)], the set of polynomials in one variable is L^p-dense in C([0,1]) and C([0,1]) is L^p-dense in L¹([0,1]) (cf. [Rud2, Theorem 2.14]).
Moreover ||f_°ⁿ || = ò₀¹ |f_°ⁿ(t)|dt = [ 1/ n!], so r(f_°) = lim_n ||f_°ⁿ ||^{[ 1/ n]} = lim_n (n!)^{[( -1)/ n]} = 0. Therefore this algebra doesn't have any character. Thus it is radical algebra.

Ref.

[Rud2] W. Rudin, Real and complex analysis, McGraw-Hill, 1986.