I have figured out how to show that the equilibrium is stable. Starting from here, Our coefficient matrix is $A = \begin{pmatrix} r - \beta I_\infty - \mu & r - \beta S_\infty \\ \beta I_\infty & \beta S_\infty - (\mu + \alpha) \end{pmatrix} = \begin{pmatrix} r - \mu - \frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} & r - \mu - \alpha \\ \frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} & 0 \end{pmatrix}$

I had to use an approach that did not rely on eigenvalues. If the determinant of the matrix is positive and of the trace of the matrix is negative, then the equilibrium is stable.

$\begin{vmatrix} r - \mu - \frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} & r - \mu - \alpha \\ \frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} & 0 \end{vmatrix} = -(r - \mu - \alpha)\frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} = (r - \mu) (\mu + \alpha) > 0$

The trace is equal to $r - \mu - \frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)}$. Using the fact that we are dealing with positive constants, $r > \mu$, and $r < \alpha + \mu$, we can show that the trace is negative.

$\begin{align}

\mu + \alpha &> \mu + \alpha - r \\[2 mm]

\frac{\mu + \alpha}{\mu + \alpha - r} &> 1\\[2 mm]

\frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} &> r - \mu

\end{align}$

Therefore, $r - \mu - \frac{(r - \mu) (\mu + \alpha)}{(\alpha + \mu - r)} < 0$. Hence our equilibrium is stable.