Propagators are independent of interactions. They only depend on the free part of the Lagrangian. For example, the KG Lagrangian reads
$$
\mathcal L_{KG}=\frac{1}{2}(\partial \phi)^2-\frac{1}{2}m^2\phi^2-\mathcal H_I(\phi)
$$
where $\mathcal H_I$ could be, eg, $\frac{g}{4!}\phi^4$. There are different definitions of the propagator (all equivalent, of course). For our purposes, the easiest definition is that it is the Green function of the KG PDE. The fastest way to find $D_F$ is to go to momentum space, and "solve" for the propagator:
$$
\mathcal L_{KG}(k)=\phi(k)(k^2+m^2)\phi(k)-\mathcal H_I(k)\equiv \phi(k) D_F^{-1}\phi(k)-\mathcal H_I(k)
$$
where we find
$$
D_F(k)=\frac{1}{k^2+m^2}
$$
(there are several conventions on the normalisation of $D_F$, so you can sometimes find some numerical factors in front of the expression above)
On the other hand, its not difficult to prove that $D_F$ agrees with your expression:
$$
D_F=\langle 0|\mathcal T\ \phi_0(x)\phi_0(y)|0\rangle
$$
where $\phi_0(x)$ is the solution of the free KG equation. If you add interactions, then the amplitude to go from $x$ to $y$ is no longer the propagator. You can call the amplitude $C(x,y)$, and we usually call this new quantity a correlator. In perturbation theory, it is not difficult to prove that $C$ can always be written in terms of the propagator (as an asymptotic expansion over some parameter that weights the interactions). The correlator depends on $\mathcal H_I$, but the propagator does not. You can prove that if $\mathcal H_I=0$, then $C(x,y)=D_F(x,y)$.
With this, we can answer your first two questions: 1) $D_F$ is not the propagation amplitude, but $C(x,y)$ is. They agree when there are no interactions. 2) The propagator is independent of interactions, but the correlator is not. The latter can be written as an asymptotic expansion where the different terms include powers of $D_F$, so that correlators can always we written in terms of propagators.
As for your third question, the propagator does depend on the PDE, as it depends on the free part of the Lagrangian. For example, if you take a different Lagrangian, as Dirac's
$$
\mathcal L_D=\bar\psi(i\not\partial-m)\psi-\mathcal H_I(\psi)
$$
then using the same line of reasoning as before we find that the propagator is
$$
S(k)=\frac{1}{\not k-m}
$$
I hope this answers your questions.
Addendum
Some people make no distinction between the concept of propagator and correlator. They call the former free propagator and the latter exact propagator (or even bare vs dressed propagators). Note that this is not very usual, for most people just call $D_F$ the propagator, and call $C$ correlator (or two-point function). IMO it is very important to understand that these are different objects, so I don't like to use the same word for them.
That being said, it seems that you just call both objects propagator. If this is the case, you should convince yourself that the free propagator is independent of interactions, and the exact one is not. So if you are going to use the same word for both, then at least you should not use the same notation. For example, in Srednicki's book he writes $\Delta$ for the free propagator, and $\boldsymbol \Delta$ for the exact one. By the way, Peskin & Schroeder do call $C$ correlator, and $D_F$ propagator (see page 82).
Anyway, if you see the links I posted, you can see that $D_F(k)=1/k^2+m^2$ (in KG's case), or $D_F(k)=1/\not k-m$ (in Dirac's). Then it is easy to see that there are no coupling constants in these expressions, so they are indeed independent of interactions. By the Dyson series you can calculate the correlators (which do depend on the coupling parameters), which are given, at fist order, by the propagators.
Best Answer
In Peskin's book he addresses your question in section 2.4. He computes $[\phi(x), \phi(y)]$ and then says $[\phi(x), \phi(y)]$ is "just a c-number so we can write: $[\phi(x), \phi(y)] = \langle 0| [\phi(x), \phi(y)] |0 \rangle$." The factor of $i$ is just a convention, I think different books use different conventions in their definitions of the propagators.