The correct answer is A.
The z-transform of a sequence $x[n]$ is defined as
$$X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}$$
The z-transform of the unit impulse sequence $\delta[n]$ is
$$X(z) = \sum_{n=-\infty}^{\infty} \delta[n] z^{-n} = 1$$
The z-transform of the shifted unit impulse sequence $\delta[n-k]$ is
$$X(z) = \sum_{n=-\infty}^{\infty} \delta[n-k] z^{-n} = z^{-k}$$
Therefore, the z-transform of the time function
$$\sum_{k=0}^{\infty} \delta[n-k] = \sum_{k=0}^{\infty} z^{-k}$$
is
$$X(z) = \frac{1}{1-z^{-1}} = \frac{z}{z-1}$$
which is option A.
Option B is the z-transform of the time function $e^{-an}$. Option C is the z-transform of the time function $e^{an}$. Option D is the z-transform of the time function $e^{-an} u[n]$.