Minimum phase systems with pole at infinity
If I see it correctly, your transfer function is

$$H(z)=\frac{1-\frac{1}{2}z^{-1}}{z^{-1}}=z-\frac{1}{2}$$

This system is not a minimum-phase system because its pole is at infinity (as you've already pointed out). You could use a trivial all-pass filter with impulse response $h[n]=\delta[n-1]$, i.e. a simple delay by one sample to change your original (non-causal) transfer function to a stable and causal transfer function

$$H_m(z)=z^{-1}H(z)=1-\frac{1}{2}z^{-1}$$

which is also minimum-phase, because it has a pole at $z=0$ and a zero at $z=1/2$. This means that all the system's poles and zeros are inside the unit circle and the minimum-phase requirement is met.