In as few words as possible, here is my question:
Why can we get the output phasor by multiplying the input phasor by the s-domain transfer function evaluated at the (complex?) frequency we're dealing with?
An example will serve to better explain:
simulate this circuit – Schematic created using CircuitLab
(Note: I am tacitly ignoring the initial conditions; for now I want to simplify the problem and only consider the steady-state response)
$$H(s)=\frac{V_C}{V_{in}}=\frac{1/(sC)}{sL+R+1/(sC)}=\frac{\frac 1{LC}}{s^2+s\frac R L +\frac 1{LC}}$$
If we wanted \$V_C\$ as a phasor, we would simply perform
$$V_C = V_{in}\cdot H(j2\pi \cdot1000)=3\angle45˚\cdot \frac{\frac 1{LC}}{-4\pi^2 \cdot 1000^2+j2\pi \cdot 1000\frac R L +\frac 1{LC}}$$
This would result in some complex number, which we would interpret as representing a sinusoid in the time domain. (Note: Already I'm confused... the input voltage waveform is not a complex number in the s-domain; it's actually some function of s).
I can understand how phasors arise naturally when solving DEs in the time domain; you assume your output is of the form \$Ae^{j2\pi ft + \phi}\$ and the time-dependence cancels out in the equation. I can also understand that multiplying by the transfer function in the s-domain produces the correct output in the time domain (provided the system is LTI). I can even understand why Ohm's Law, KVL, and KCL work in the s-domain.
However, after all that I can't get my head around this "abuse of notation". Phasors and s-domain expressions shouldn't have any business hanging around each other! So what am I missing here?