Why do phasors work with s-domain transfer functions?
Actually, phasors and s-domain expression do have some relationship. Recall that the 's' variable in the laplace transform is defined as: 

$$ s = \sigma + \text{j}\omega$$

So, when you substitute \$s\$ by \$\text{j}\omega\$ in a transfer function, you are taking your function to the phasor domain (which will produce the steady-state solution only, not the transient response).

Now, remember that when using phasors, you're taking advantage of Euler's identity, that is, \$e^{\text{j}\omega t}=\cos(\omega t)+\text{j}\sin(\omega t)\$. Even though you have a real source, say it is \$v(t)=\text{V}_o\cos(\omega t)\$, you can use Euler's identity to express it as 

$$ v(t)=\text{V}_oe^{\text{j}\omega t}$$

or more rigorously defined as 

$$ v(t)=\Re{[\text{V}_oe^{\text{j}\omega t}]}$$

where \$\Re\$ means that you want the real part of the expression. That's the case since your source is a cosine, or the real part of Euler's identity. 

Alternatively, if you source were a sine, then

$$ v(t)=\Im{[\text{V}_oe^{\text{j}\omega t}]}$$

where \$\Im\$ means that you want the imaginary part of Euler's identity.

That means, that once you solve your circuit, you need to take either the real (if your source was a cosine) or the imaginary (if your source was a sine) part of the complex valued solution.

Hope it helps!