The Gel'fand spectrum of $\ell^1 \mathbb N$ is indeed the closed unit disc. After all, every functional to $\mathbb C$, must be given by sending the generator to some complex number. It is easy to see, that this works if and only if this complex number lies in the unit disc.

Your second question is rather nice. Anyhow, I think that there cannot be any commutative example (at least if it embeds into the algebra of continuous functions on the Gel'fand spectrum). As soon as there is the unit in the algebra of functions on the Gel'fand spectrum, then the Banach algebra contains at least an invertible element, and hence also the unit.

However, and now it is getting more interesting: there are non-unital Banach algebras whose universal $C^\star$-algebra has a unit. (Note that this is precisely the non-commutative analogue of your question in Gel'fand theory.)
Indeed, consider a group $\Gamma$ with Kazhdan's property (T), its $\ell^1$-algebra and the augmentation ideal $\omega(\Gamma) \subset \ell^1 \Gamma$. It is well-known that $C^\star(\Gamma)$ splits of the unit as a direct summand. The remaining summand is the universal $C^*$-algebra of $\omega(\Gamma)$ and it is unital.