Why Z-transform is considered as separate transform?
There is indeed a transform called discrete Laplace transform and it is of course closely related to the $\mathcal{Z}$-transform. The (unilateral) discrete Laplace transform of a sequence $f_n$ is defined by (cf. link)

$$\mathcal{L}_T\{f_n\}=\sum_{n=0}^{\infty}f_ne^{-snT}$$

with some $T&gt;0$. The discrete Laplace transform can be interpreted as the Laplace transform of a sampled function $f(t)\cdot\sum_n\delta(t-nT)$  with $f_n=f(nT)$.

In practice it is not convenient to have the factor $e^{sT}$ appear everywhere. If one substitutes $z=e^{sT}$, the discrete Laplace transform is called (unilateral) $\mathcal{Z}$-transform:

$$\mathcal{Z}\{f_n\}=\sum_{n=0}^{\infty}f_nz^{-n}$$

The same can obviously be done for the bilateral versions of the transforms, where the integrals and the sums start at $-\infty$.