The Laplace transform of $e^{-at}$ is $G(s)=\frac{1}{s+a}$, and the Z-Transform of a discretely sampled version of the time domain function is $G(z)=\frac{z}{z-e^{-aT}}$.
When you multiple $G(s)$ by the laplace transform of a step input $1/s$ the result is $Y(s)=\frac{1}{s(s+a)}$, when $G(z)$ is multiplied by the z-domain step input $\frac{z}{z-1}$, the result is $\frac{z^2}{(z-1)(z-e^{-aT})}$.
This does not equal the Z transform that corresponds to the same time domain function as $Y(s)$, which is:
$$\frac{z(1-e^{-aT})}{a(z-1)(z-e^{-aT})}$$
Therefore, if you treat these transforms as transfer functions they have different step responses. Why is that?