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Clearly all eigenvalues apart from the eigenvalue $\lambda(\epsilon)$ with $\lambda(0) = 1$ stay in the open unit disk for $\epsilon$ sufficiently small. To see what happens to the last eigenvalue, use eigenvalue first-order perturbation theory, for instance Theorem 1 in https://arxiv.org/abs/1903.00785 : if $v$ and $w^*$ are the right and left eigenvalue associated to $\lambda(0)$, then $$ \frac{d\lambda(\epsilon)}{d\epsilon} = \frac{w^*(-I)v}{w^*v} = -1. $$ As the derivative is negative, for a sufficiently small $\epsilon > 0$ we have that $\lambda(\epsilon)$ is in the unit disk.