Discrete time Final Value Theorem applied to feedback system
I would derive the total transfer function directly in the transform domain. Your input-output equation can be written as
$$Y(z)\big(1-z^{-1}\big)=\alpha G(z)z^{-1}Y(z)+\beta z^{-1}X(z)\tag{1}$$
where $G(z)$ is the transfer function of the high-pass filter. From $(1)$ you directly obtain the transfer function
$$H(z)=\frac{Y(z)}{X(z)}=\frac{\beta z^{-1}}{1-z^{-1}\big(1+\alpha G(z)\big)}\tag{2}$$
which clearly has a pole at $z=1$ because $G(1)=0$.

The mistake in your calculation is in step 2. You increased the lower bound of the summation from $1$ to $2$ without changing the power of $\phi$ from $i$ to $i-1$. In the transfer function this results in a $\phi^2$ in the last part of the denominator instead of a $\phi$. The denominator should be:
$$D(z)=1 - (1+\Psi\phi)z^{-1}-\Psi(\phi-1)\frac{\phi z^{-2}}{1-\phi z^{-1}}\tag{3}$$
For $z=1$ this evaluates to
$$D(1)=1 - (1+\Psi\phi)-\Psi(\phi-1)\frac{\phi }{1-\phi }=1 - (1+\Psi\phi)+\Psi\phi=0\tag{4}$$