The Banach space considered is the following: $(l^{\infty}(\mathbb{Z}), \|\cdot\|_{*})$ with $\|x\|_{*}=\|(...,x_{-1},x_{0},x_{1},...)\|_{*}=|x_{0}|+\text{sup}_{k\neq 0}|x_{k}|$.
Define $A$, an operator on this space as follows:
$(Ax)_{k}=\alpha_{k}x_{k+1}$ with $\alpha_{k}=1$ if $k\neq 0$ and $\alpha_{0}=\frac{1}{3}$.
It can be shown that $A$ is invertible and $\|A\|\leq 1, \|A^{-1}\|\leq 3$, and hence $\sigma(A)\subseteq \{\lambda\in \mathbb{C}:\frac{1}{3}\leq|\lambda|\leq 1\}$. It can be shown that $1\in \sigma_{p}(A)$, i.e., the point spectrum of $A$, with eigenvector $x$ given by $x_{k}=1$ if $k\leq0$ and $x_{k}=3$ if $k\geq 1$. This can be extended to show that, in fact, the entire unit circle $\mathbb{T}$ is contained in the point spectrum, and hence the spectrum of $A$.
Can it be shown that the spectrum of $A$ is, in fact, equal to the unit circle?
I would be grateful for a hint on how to proceed.
$\textbf{Edit:}$ I wonder if the following argument is correct:
It can be seen that $\|(A^{-1})^{n}\|=3 \, \forall n$. By the spectral radius formula, it follows that $r(A^{-1})=1$, where $r(A^{-1})$ is the spectral radius of $A^{-1}$. This implies that $\sigma(A)\subseteq \mathbb{T}$, and so in fact $\sigma(A)=\mathbb{T}$.
To prove what you want, it suffices to prove that if $|\lambda| \neq 1$, then $(A-\lambda)$ is bijective. I was able to prove injectivity - surjectivity seems hard to just brute force, so I will post this and let someone else come up with an elegant answer :)
Injectivity: Suppose $Ax = \lambda x$, and $\lambda, x\neq 0$, then $$ \frac{1}{3}x_1 = \lambda x_0 \text{ and } x_{k+1} = \lambda x_k \quad\forall k\neq 0 $$ Hence $x_0 \neq 0$, so assume without loss of generality that $x_0 =1/3$, so that $$ x_1 = \lambda $$ Now by induction, one can see that $x_n = \lambda^n$ for all $n \in \mathbb{Z}$. But $x \in \ell^{\infty}(\mathbb{Z})$, so it follows that $|\lambda| = 1$.
In other words, by what you have said, $\sigma_p(A) = \mathbb{T}$.