I believe (?) they are similar concepts; a delay in a difference equation is considered synonymous to a derivative in a differential equation. But
$$ f'(t) \rightarrow sF(s) - f(0) $$
$$f[k-1] \rightarrow zF(z)$$
This continues for higher derivatives/greater delays. Is there a reason for the non-existent initial value term in the z-transform?
It comes from the fact that Laplace transforms are usually unilateral (one-sided), while Z-transforms are bilateral (two-sided). Indeed, one has $$ \mathscr{L}[f'(t)](s) = \int_0^\infty f'(t)e^{-st}\,\mathrm{d}t = \left.f(t)e^{-st}\right|_0^\infty + s\int_0^\infty f(t)e^{-st}\,\mathrm{d}t = sF(s) - f(0), $$ by integration by parts, but $$ \mathcal{Z}[f(k\color{red}{+}1)](z) = \sum_{k\in\mathbb{Z}}f(k+1)z^{-k} = \sum_{k\in\mathbb{Z}}f(k)z^{-(k-1)} = zF(z). $$ If, on the contrary, we consider the two-sided Laplace transform and the one-sided Z-transform, the same expressions become $$ \mathscr{L}[f'(t)](s) = \int_{-\infty}^\infty f'(t)e^{-st}\,\mathrm{d}t = \left.f(t)e^{-st}\right|_{-\infty}^\infty + s\int_{-\infty}^\infty f(t)e^{-st}\,\mathrm{d}t = sF(s) $$ $-$ with appropriate conditions on the asymptotic behaviour of $f$ $-$ and $$ \mathcal{Z}[f(k+1)](z) = \sum_{n\ge0}f(k+1)z^{-k} = f(1) + \sum_{n\ge0}f(k)z^{-(k-1)} = zF(z) + f(1). $$ In conlusion, unilateral transforms produce boundary terms.
Addendum. In order to avoid this problem but keep the one-sided Laplace transform at the same time (because its two-sided version often has convergence issues), some people work implicitly with $f(t)u(t)$ instead of $f(t)$, where $u(t)$ is the Heaviside function, and modify a little bit the definition of the Laplace transform, as follows :
$$ F(s) := \int_{\color{red}{0^\mathbf{-}}}^\infty f(t)e^{-st}\,\mathrm{d}t \quad\Rightarrow\quad \mathscr{L}[f'(t)](s) = sF(s), $$ since boundary terms are killed by the fact that $u(0^-) = 0$.