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The first algebra exists. Let $A=L^1(0,1)$ with truncated convolutionThe Tonelli theorem givesCommutativity follows by replacing $s$ with $t-s$; associativity follows from the Fubini theorem applied to the simplex $s,u\ge 0$, $s+u\le t$. Together with completeness of $L^1$, this makes $A$ a commutative Banach algebra, not necessarily unital. Adjoin an identity using unitization:This norm is complete and submultiplicative by the preceding bound, and $e=(1,0)$ is the identity. This is the Volterra convolution algebra on a finite interval with an identity adjoined.

Take $x=(0,1)$, where $1(t)=1$. Direct induction under convolution givesThus no power vanishes. For every $\lambda \ne 0$ the seriesconverges absolutely, since its norms have factorial denominators. Multiplication by $\lambda e-x$ telescopes to $e$, so it is an inverse. On the other hand $x$ cannot be invertible: its scalar coordinate is zero, and scalar coordinates multiply, so $xy$ can never equal $e$. HenceThis explicitly exhibits a nonnilpotent quasinilpotent element without assuming a spectral-radius formula.

The second algebra does not exist. Suppose it did, and choose a nonscalar $y$. Nonemptiness of its spectrum of an element, together with $\rho (y)=0$, gives ${\sigma }_B(y)=\{0\}$. But $y+e$ is still nonscalar: if $y+e=\alpha e$, then $y=(\alpha -1)e$. By translation of the spectrum by a scalar,This contradicts the required zero spectral radius for every nonscalar element. The contradiction already uses the spectral condition alone; the nonvanishing-power condition cannot remedy it.