Z transform of a function with delay?
In my opinion, the most straightforward way to solve this problem is to consider the time domain signal as it passes through the signal chain. A unit step at the input when sampled and put through a ZOH becomes just another continuous-time unit step. Consequently, the Laplace transform of the continuous-time signal just before the sampler at the output is given by

$$C(s)=\frac{e^{-s}}{s(s+a)}=\frac{e^{-s}}{a}\left[\frac{1}{s}-\frac{1}{s+a}\right]\tag{1}$$

which corresponds to the time domain signal

$$c(t)=\frac{1}{a}\left[1-e^{-a(t-1)}\right]u(t-1)\tag{2}$$

Sampling $c(t)$ at $t=kT$ gives

$$c(kT)=\frac{1}{a}\left[1-e^{-a(kT-1)}\right]u(kT-1)\tag{3}$$