Yes, that is what I meant. But now I am confused too. What I was trying to say was this:
As a function, the input of H(s) can only be a complex number, and its output can only be a complex number. H(s) itself is describing the input-output ratio of damped sinusoids.
Suppose x(t) = etsin(t), and you run it through a system that has the Laplace transform H(s) = s. Then in this case, since 's' is the combination of the exponent and the frequency, s = 1 + i. H(s) evaluated at s=1+i is 1+i. The magnitude is sqrt(2), the phase is 45o. This means that the magnitude of x(t) will be modified by sqrt(2) and its phase would change by 45o. So
y(t) = sqrt(2) et sin(t +45o).
In this example, there is no need to do Laplace transform of x(t), multiply it to H(s), and do inverse Laplace transform of Y(s) to get y(t), because this is done using the meaning of H(s), and you get y(t) by reading the gain and phase shift.
I don't know how to explain the case with unit step function. It works but there seems to be no direct meaning while the unit step input is respresented by 1/s in s-domain. It is like,
"Let's treat the unit step input as if it is part of the system, and that the input to this arbitrary system is some imaginary damped sinusoid described by the shorthand complex notation s; then we observe that the system is composed of linear operators that can be multiplied together, and use inverse Laplace transform to find the overall impulse response--which is really the output y(t) we wanted to find originally. We do it like this because the math becomes easier, not because there is a direct meaning."In my way of seeing it, the Laplace transform of signal x(t) = eatsin(wt)u(t), which is X(s) = w/( (s-a)2+w2 ) is not really a representation of the signal x(t), but a representation of a system with impulse response x(t).