How do causality and stability affect the response of a linear time-invariant system?
Technically, the Laplace transform already implies a causal system by its definition:
$$F(s)=\int_0^\infty f(t)e^{-st}dt$$
As you can see, the integral starts from 0, so this implies that \$f(t &lt; 0) = 0\$. In other words, a non-causal system would react before it received the Dirac pulse or \$f(t &lt; 0) \neq 0\$. A regular Laplace transform would not work in this case.
If the system is causal, the method of finding \$h(t)\$ does not change whether or not the system is stable or unstable. The example you gave has a pole in both left- and right-half plane and is therefore unstable, but the inverse Laplace transform can still be calculated (IIRC) to be
$$\frac{e^{2t}}{3} - \frac{e^{-t}}{3}$$
As you can see, the first term explodes as \$t\$ increases, hence the system is unstable, but I didn't have to use any weird techniques or something.
If you are given the information that the transfer function is a two-sided Laplace transform, then you cannot assume anymore that the system is causal while you might still reconstruct the impulse response from it, but I have never encountered such a situation.