How is causality in Laplace transform related to Fourier transform?

If we are given a function $H(s)$ and we're told that it is a Laplace transform, then there are usually many possible corresponding time-domain functions. Let's assume that $H(s)$ is rational (so we can talk about poles), then the possible regions of convergence (ROCs) are limited by the real parts of the poles. If the ROC is to the right of the rightmost pole, the corresponding time-domain function is right-sided, if the ROC is to the left of the leftmost pole, the time-domain function is left-sided, and if the ROC is between two poles, we have a two-sided time-domain function.

Let's now assume that there are no poles on the imaginary axis. If we set $s=j\omega$ we obtain a valid Fourier transform $H(j\omega)$. Its inverse Fourier transform equals the time domain function corresponding to the (unique) ROC of $H(s)$ which includes the imaginary axis.

Example:

$$H(s)=\frac{2a}{a^2-s^2}=\frac{1}{s+a}-\frac{1}{s-a},\qquad a>0\tag{1}$$

Depending on the ROC, $H(s)$ is the Laplace transform of the following three time-domain functions:

\begin{align*} h_1(t) &= \left[e^{-at}-e^{at}\right]u(t),& \textrm{Re}(s)>a \\ h_2(t) &= \left[e^{at}-e^{-at}\right]u(-t),& \textrm{Re}(s)<-a \\ h_3(t) &= e^{-at}u(t)+e^{at}u(-t)=e^{-a|t|},& -a<\textrm{Re}(s)<a \end{align*}

The inverse Fourier transform of

$$H(j\omega)=\frac{2a}{a^2+\omega^2}\tag{2}$$

is given by $h_3(t)$, which corresponds to the only stable impulse response, i.e., to the ROC which includes the imaginary axis. The functions $h_1(t)$ and $h_2(t)$ don't have a Fourier transform.

Analytic continuation of $H(s)$ to all points of the complex plane except for the poles does not affect any of the above, and it is not related to the Fourier transform.