Let $E$ be a Banach space and $$H_p=\{f\in\mathcal L(E): f(v)=\lambda v\ (v \neq 0)\implies |\lambda|\neq 1\}$$ be the set of linear operators which don't have an eigenvalue with norm $1$. Is $H_p$ dense in $\mathcal L(E)$?
This is false if we consider the set of hyperbolic transformations $$H=\{f\in\mathcal L(E):\lambda \in \sigma(f)\implies |\lambda|\neq 1\}$$ as José Alves shows in these lecture notes, on example 5.3.
The space $H_p$ feels much bigger than $H$, plus the proof that the hyperbolic matrices in a finite dimensional space are dense by perturbing the eigenvalues makes me think that this is true.