The complex exponential e^(st) is an eigenfunction of Linear time invariant (LTI) systems. This implies that if the input is a complex exponential e^(st), the output will be the same complex exponential scaled by a scalar. The value of the scalar can be found from the Laplace transform of the impulse response of the system. Just to verify this I tried to find the output of the system by the method of convolution. The output obtained through convolution is not a signal complex exponential. It's a totally different answer obtained from eigenfunction property. Why am I getting different answers?
If it converges : $$e^{st} \longrightarrow \boxed{\ h(t)\ } \longrightarrow e^{st}\ast h(t) = \int_{-\infty}^\infty h(\tau)e^{s (t-\tau)}d\tau = e^{st} H(s)$$ where $H(s) = \int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau$ is the (bilateral) Laplace transform of $h(t)$