When inverting a transfer function, solving for the input using the output does the causality status change
As you've pointed out, inversion leads to poles at locations of the zeros of the original transfer function and vice versa. Assuming that $G(z)$ is causal and stable (i.e., it has all its poles inside the unit circle), we have to distinguish $3$ cases:


$G(z)$ has at least some zeros outside the unit circle. This means its inverse has some poles outside the unit circle, and consequently, it cannot be causal and stable. If $G(z)$ has no zeros on the unit circle, there exists a stable impulse response corresponding to $1/G(z)$ but it cannot be causal. This is because a transfer function does not uniquely determine an impulse response. We can get different impulse responses corresponding to different regions of convergence of $1/G(z)$.
$G(z)$ has some of its zeros on the unit circle. No stable inverse exists because $1/G(z)$ has poles on the unit circle.
$G(z)$ has all its zeros inside the unit circle, i.e., it is a minimum-phase system. Consequently, $1/G(z)$ also has all its poles and zeros inside the unit circle (i.e., it is minimum-phase), and can be implemented by a causal and stable system.