Z transform of a function with delay? i have this open loop system , and i've been asked to find out the response $C(kT)$ due to a unit step input.

I am able to find the transfer function without the delay unit i.e $$\frac{C(z)}{R(z)}=\left(1-z^{-1}\right)\cdot \mathcal{Z}\left\{\frac{1}{s(s+a)}\right\}$$
i can find this ZT easily and find $C(kT)$

But how to deal with the exponential term i.e $e^{-s}$ ? then my pulse transfer function becomes 
$$\frac{C(z)}{R(z)}=\left(1-z^{-1}\right)\cdot \mathcal{Z}\left\{\frac{e^{-s}}{s(s+a)}\right\}$$
Can i also consider it as a delay ? and take it out of the braces ? i.e
$$\frac{C(z)}{R(z)}=\left(1-z^{-1}\right)\cdot z^{-1} \cdot \mathcal{Z}\left\{\frac{1}{s(s+a)}\right\}$$
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

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