Notation:
Say we have a continuous-time signal $x(t)$. We can take its Fourier transform and get $$x(t) \longleftrightarrow X(f)$$ We know that multiplying two signals in the time domain corresponds to convolving their frequency domain representations. We also know that a Dirac comb in the time domain is a Dirac comb in the frequency domain, so we can define $x_s(t)$ (the sampled version of $x(t)$): $$x_s(t) \equiv x(t)\cdot\mathrm{comb}(t)$$ $$x_s(t) \longleftrightarrow X_s(f)=X(f)\star \mathrm{comb}(f)$$
Where I have used the $\star$ operator to denote convolution.
We're almost done setting up notation. We'll finish off with the Laplace- and Z-transforms: $$x(t)\longleftrightarrow \mathcal{X}(s)$$ and, using the sifting property, $$x_s(t)\longleftrightarrow \mathcal{X}_s(s)=\int_{-\infty}^{\infty}x(t)\cdot\mathrm{comb}(t)\cdot e^{-st}\ dt=\sum_{n=-\infty}^{\infty}x[n]e^{-sn}$$
where I have hand-waved away the spacing between the impulses of the comb function.
Finally, we define $z\equiv e^{s}$ and get $$x_s(t)\longleftrightarrow\mathcal{X}_z(z)=\sum_{n=-\infty}^{\infty}x[n]z^{-n}$$
the Z-transform of $x[n]$.I've done away with the spacing between the impulses for the sake of simplicity.
The question:
The effects of sampling in the time domain cause a "periodicization" in the frequency domain (i.e. the convolution with a comb function). If your sampling rate is high enough, then you can completely recover $x(t)\longleftrightarrow X(f)$ from $X_s(f)$ by multiplying $X_s(f)$ with a rect function, in order to cancel out the "copies" of the spectrum produced in the convolution.
What I would like to know is if there is a strategy to at least approximate the Laplace transform of $x(t)$ knowing its Z-transform, $\mathcal{X}_z(z)$. I have tried to work it out by investigating the effects of sampling:
$$x(t)\cdot \mathrm{comb}(t) \longleftrightarrow \mathcal{X}(s)\star ??$$
In fact, the Laplace transform of $\mathrm{comb}(t)$ is already a bit strange:
$$\mathscr{L}\{\mathrm{comb}(t)\}=\int_{-\infty}^{\infty}\sum_{k=-\infty}^{\infty}\delta(t-k)\cdot e^{-st}\ dt $$ $$= \sum_{k=-\infty}^{\infty}e^{-sk}= \sum_{k=0}^{\infty}e^{-sk} + \sum_{k=1}^{\infty}e^{sk}$$
(Note the negation of the exponent in the second summation on the right hand side). Problem is, I think this sum should diverge. It's not even clear what the Laplace transform of the comb function is, so I really have no idea what the convolution would produce in the s-domain, let alone how to undo it.
Anyone have any insight here?
