Let $A=\ell^1 (\mathbb Z)$. I read that it is possible to identify $S^1$ with the character space $ \Omega (A)$. But I have constructed a proof that identifies $ \Omega (A)$ with $\mathbb D$, the closed unit disk. Shouldn't this be a contradiction?
My proof is the following:
Define a map $\varphi$ by defining for $\lambda \in \mathbb D$: $$ \varphi(\lambda) = \tau_\lambda \hspace{0.2cm} \text{ where } \hspace{0.2cm} \tau_\lambda (f) = \sum_{n \in \mathbb Z} f(n) \lambda^n$$
We claim that $\varphi$ is a bijective map $\mathbb D \to \Omega (\ell^1(\mathbb Z))$: That it is a well-defined map into $\Omega (\ell^1(\mathbb Z))$ is clear since for $\lambda \in \mathbb D$ the series $ \sum_{n \in \mathbb Z} f(n) \lambda^n$ converges absolutely and hence $\tau_\lambda$ is additive, multiplicative and bounded.
Next we argue that $\varphi$ is surjective. To this end let $\tau \in \Omega (\ell^1 (\mathbb Z))$ and let $\lambda = \tau (\chi_{\{1\}})$. We claim that $\lambda \in \mathbb D$: Certainly, $|\lambda| = |\tau (\chi_{\{1\}})| \le \|\tau\| \le 1$. Next we claim that $\tau = \tau_\lambda$: But this is also clear. The set $\{\chi_{\{n\}}\}_{n \in \mathbb Z}$ generates $\ell^1 (\mathbb Z)$ as an algebra and for all $n \in \mathbb Z$: $\tau(\chi_{\{n\}}) = \lambda^n = \tau_\lambda (\chi_{\{n\}}) $ hence $\tau = \tau_\lambda$.
We conclude the proof by showing that $\varphi$ is injective. To this end, let $\lambda, \lambda' \in \mathbb D$ be such that $\tau_\lambda = \tau_{\lambda'}$. Then in particular, $\lambda = \tau_\lambda (\chi_{\{1\}}) = \tau_{\lambda'} (\chi_{\{1\}}) = \lambda'$.
Where is the mistake in my proof?
Apparently you forgot about the negative powers. If for example we consider $a \in \ell^1(\mathbb{Z})$ given by $a_n = 2^{-\lvert n\rvert}$, the corresponding Laurent series
$$f(z) = \sum_{n=-\infty}^{+\infty} a_n z^n = \sum_{n=1}^\infty a_{-n}z^{-n} + \sum_{n=0}^\infty a_nz^n = \sum_{n=1}^\infty \frac{1}{(2z)^n} + \sum_{n=0}^\infty \left(\frac{z}{2}\right)^n = \frac{1}{2z-1} + \frac{2}{2-z}$$
converges only in the annulus $\frac{1}{2} < \lvert z\rvert < 2$. For $b_n = \frac{1}{1+n^2}$, the series converges only for $\lvert z\rvert = 1$.
To determine $\Omega(A)$, consider the powers of $\chi_{\{1\}}$. For any $k \in\mathbb{Z}$, we have $\chi_{\{1\}}^k = \chi_{\{k\}}$, thus $\lVert\chi_{\{1\}}^k\rVert = 1$ for all $k\in\mathbb{Z}$. So if $\varphi \in \Omega(A)$ and $\lambda = \varphi(\chi_{\{1\}})$, then
$$\lvert \lambda^k\rvert \leqslant \lVert \chi_{\{1\}}^k\rVert = 1$$
for all $k \in \mathbb{Z}$. Choosing $k = 1$ yields $\lvert\lambda\rvert \leqslant 1$, and $k = -1$ yields $\lvert \lambda\rvert \geqslant 1$, hence $\lvert\lambda\rvert = 1$. It is then clear that $\varphi$ is the evaluation homomorphism in $\lambda\in S^1$, and conversely, every $z\in S^1$ corresponds to an evaluation homomorphism. We have thus a bijection between $S^1$ and $\Omega(A)$. It is easy to verify that $z \mapsto \varphi_z$ is continuous, and since both spaces are compact (Hausdorff) spaces, it follows that the bijection is a homeomorphism.