ROC of the product of two Z-Transforms
Note that for any impulse response $h[n]$, the step response is given by the cumulative sum of $h[n]$:

$$a[n]=\sum_{k=-\infty}^nh[k]\tag{1}$$

Its $\mathcal{Z}$-transform is then

$$A(z)=\sum_{n=-\infty}^{\infty}a[n]z^{-n}=\sum_{n=-\infty}^{\infty}z^{-n}\sum_{k=-\infty}^nh[k]\tag{2}$$

The region of convergence (ROC) of $(2)$ is the range of values of $|z|$ for which the sum in $(2)$ converges.

For your example, you have three different impulse responses with the same expression for their $\mathcal{Z}$-transform, each with a different ROC. You are right that the ROC of $A(z)$ is the intersection of the ROC of $h[n]$ and the ROC of the unit step $u[n]$.

For the ROC $|z|&lt;\frac12$ you get an empty intersection. This simply means that the sum $(2)$ converges nowhere in the $z$-plane, i.e. the $\mathcal{Z}$-transform of the step response $a[n]$ doesn't exist in this case. The reason is that the corresponding impulse response is anti-causal and exponentially increasing towards negative values of $n$, so from $(1)$, $a[n]$ is infinite for any value of $n$.

For the ROC $\frac12&lt;|z|&lt;2$ you have the intersection $1&lt;|z|&lt;2$, i.e., the step response $a[n]$ is a two-sided sequence with a $\mathcal{Z}$-transform $A(z)$ that converges in that range of values of $|z|$.

For the ROC $|z|&gt;2$ the intersection is $|z|&gt;2$, and, consequently, the step response $a[n]$ is an exponentially increasing right-sided sequence with a $\mathcal{Z}$-transform $A(z)$ that converges for $|z|&gt;2$.