Relationship between Dyson equations from different problems Recently, I noticed that the Dyson equation
$$G=G_0+G_0\Sigma G$$
is used not only in quantum field theory but in some other branches of physics. For instance:
1. Wave equation
From the wave equation in a random media
$$\mathcal{L}G(x,x_0)\equiv \left[\Delta_x+k^2 (1+\epsilon(x))\right]G(x,x_0)=\delta(x-x_0)$$
and
$$\mathcal{L}_0G_0(x,x_0)\equiv\left[\Delta_x+k^2 \right]G_0(x,x_0)=\delta(x-x_0)$$
, we get
$$\mathcal{L}_0G(x,x_0)=-k^2\epsilon(x)G(x,x_0)+\delta(x-x_0)$$
$$G(x,x_0)=\int dx_1\,\,G_0(x,x_1)\mathcal{L}_0G(x_1,x_0)=G_0(x,x_0)-\int dx_1\,G_0(x,x_1)k^2\epsilon(x_1)G(x_1,x_0)$$
, which obviously has the form of the Dyson equation.
2. Single-particle propagator - 1
Let $U(t,t_0)$ be the time evolution operator. The single particle propagator
$$K(xt;x't')=\langle x|U(t,t_0)|x'\rangle$$
satisfies follwoing equations:
$$\left[i\hbar\partial_t -\mathcal{H}\right]K(xt;x't')=\delta(x-x')\delta(t-t')$$
$$\left[i\hbar\partial_t -\mathcal{H}_0\right]K_0(xt;x't')=\delta(x-x')\delta(t-t')$$
where $\mathcal{H}=\mathcal{H}_0+V$. Those equations are mathematically identical to the wave equations above, i.e. $G\leftrightarrow K,\,-k^2\epsilon\leftrightarrow V$.
3. Single-particle propagator - 2
Using the Dyson series for $U$ in the interaction representation,
$$U=1+\sum_{n=1}^{\infty}\frac{1}{n!}\left(-\frac{i}{\hbar}\right)^n\int\int\cdots \mathcal{T}\left[V(\tau_1)\cdots V(\tau_n)\right]$$
, we can also get the Dyson equation; strictly speaking, we can get the Dyson expansion of $K$ if we expand the Dyson equation in an iterative manner: $K=K_0+K_0\sum(K_0+K_0\sum(\cdots))$.
4. Green function in condensed matter physics
At zero temperature, the casual Green function is
$$iG_{\alpha \alpha'}=\frac{\langle \psi_0|\mathcal{T}U(\infty,-\infty) \Psi_{\alpha}(x,t)\Psi_{\alpha'}^{\dagger}(x',t')|\psi_0 \rangle}{\langle \psi_0 |U(\infty,-\infty)|\psi_0 \rangle }$$
and it is not a Green function in a mathematical manner, i.e. $\left[i\hbar\partial_t -\mathcal{H}_0\right] iG_{\alpha \alpha'} \neq \delta$. Indeed, the unperturbed Green function $iG_{0,\alpha \alpha'}$ is a mathematical Green function. Using the Dyson expansion for $U$ in the interaction picture and the Wick's theorem, we can express $iG_{\alpha \alpha'}$ in terms of $V$ and $iG_{0,\alpha \alpha'}$. Such expression is known to satisfy the Dyson equation: $G=G_0+G_0\sum(G_0+G_0\sum(\cdots))$.
Question
Obviously, the case 1 and 2 are closely related to each other. The case 3 and 4 are connected, either. However, I am not sure if there is a mathematical explanation about the reason that the Dyson expansion (case 3 & 4) leads to the result identical to the simple algebra (case 1 & 2).
I think the fact that the Dyson expansion produces not the Dyson equation itself $G=G_0+G_0\Sigma G$ but its iterative version $G=G_0+G_0\sum(G_0+G_0\sum(\cdots))$ would be a clue; but I have not made any progress so far.
 A: Green's functions and example 1
First off, you are using the symbol $G$ and $K$ in 1 & 2 for Green's functions.
So let's define a Green's function first as $G(t,u)$ such as:
$$ \mathcal{L}\,G(t,u) = \delta(t-u), $$
where $\mathcal{L}$ is a linear differential operator governing the evolution of the system.
The Green's function is used in solving the dynamics $x(t)$ caused by a source $f(u)$ by integrating over it:
$$\mathcal{L}\,x(t) = \int \mathrm{d}u\,\mathcal{L}\,G(t,u)\, f(u) = \int \mathrm{d}u \ \delta(t-u)f(u) = f(t).$$
These are widely used in classical physics, such as in example 1.
Example 2 & Quantum mechanics
Quantum mechanics (first quantisation) is described by the Schrödinger equation, i.e. just a linear differential operator $\mathcal{L}$ like above. Similarly, then, one has:
$$ \phi(x,t_x) = \int \mathrm{d}y\, G^+(x,t_x,y,t_y) \phi(y,t_y), $$
where the Green's function $G^+$ propagates the particle, described by a wavefunction $\phi$, from position $y$ and time $t_y$ to a position $x$ at time $t_x$. The $+$ sign is to prevent particles from travelling back in time, so technically $G^+ = \theta(t_x-t_y)G$ is the retarted Green's function.
Hence the connection between a Green's function and a propagator in quantum mechanics. Generally, the propagator $G^+$ can be written:
$$ G^+(x,t_x,y,t_y) = \theta(t_x-t_y)\, \langle x(t_x) | y(t_y)\rangle, $$
i.e. the probability amplitude that a particle in state $|y\rangle$ at time $t_y$ ends ip in state $|x\rangle$ at time $t_x$.
The fact that the propagator is non-zero for causally disconnected events, i.e. events that are space-like separated, is the failure of first quantisation and the need of second quantisation (quantum field theory)
Dyson's equation
Green's function in the energy domain can be written as:
$$G^+(x,y,E) = \sum_n \frac{i\phi_n(x) \, \phi_n^\ast (y)}{E-E_n} \propto \frac{1}{E-H_0}. $$
easily justified by taking the Laplace equation for the electric potential $\phi$ in vacuo $\nabla^2\phi = 0$. The Green's function solution is $\epsilon_0 \nabla^2V(\mathbf{r}) = - \delta^{(3)}(\mathbf{r}-\mathbf{r}_0)$, to which you know the solution to be a point charge with $V(\mathbf{r}) = 1/|\mathbf{r}-\mathbf{r}_0|$
Green's functions allow us to interpret a perturbation problem in terms of a propagating particle. A perturbation interrupts the propagation via a scattering process, following which free-particle propagation resumes.
A generic Hamiltonian $H = H_0 + V$ is decomposed in to its solvable, free-particle propagator-giving solution $H_0$ and in a non-analytically solvable perturbation $V$.
The Green's function is given by $$G \propto \frac{1}{E-H} = \frac{1}{E-H_0 -V},$$ which would be called the full propagator to distinguish it from the free propagator $G_0 \propto 1/(E-H_0)$. 
In a perturbative approach, where $V \ll H_0$, the expression above can be written as:
$$ G = \frac{1}{E-H_0-V} = \frac{1}{E-H_0} + \frac{1}{E-H_0}V\frac{1}{E-V_0}+\frac{1}{E-H_0}V\frac{1}{E-V_0}V\frac{1}{E-V_0} + \dots $$ or $$G = G_0 + G_0 V G_0 + G_0 V G_0 V G_0 + \dots $$
which is a geometric series, rewritten as:
$$ G = G_0(1 + VG_0 + VG_0VG_0 + \dots) = \frac{G_0}{1-V G_0} = \frac{1}{G_0^{-1}-V},$$
which is known as Dyson's equation.
Example 3
The Schrödinger equation for a generic Hamiltonian $H$ is:
$$ i\hbar \frac{\mathrm{d}}{\mathrm{d}}|\psi_t\rangle = H |\psi_{t_0}\rangle.$$
$$ \implies |\psi_t\rangle = U(t,t_0) |\psi_{t_0}\rangle, $$
where $U$ is the time-evolution operator.
In the interaction picture, the Hamiltonian is split between its free part $H_0$ and interacting part $H'$ such that the time-evolution of the interaction picture Haimltonian $H_I$ is given by $H_I(t) = e^{iH_0 t/\hbar}H'e^{-iH_0t/\hbar}$.
The time evolution operator in this picture then becomes $U_I = e^{iH_0t/\hbar}Ue^{-iH_0t/\hbar}, and satisfies the following equation of motion:
$$ i\hbar \frac{\mathrm{d}}{\mathrm{d}t}U_I = H_I U_I. $$
The solution to the above is :
$$ U_I(t,t_0) = 1 - \frac{i}{\hbar} \int_{t_0}^t \mathrm{d}t' H_I(t', t_0)\,U_I(t', t_0) ,$$
and it can be iterated by keeping plugging in the expression for the $U_I$ in the integral:
$$ U_I(t,t_0) = 1 - \frac{i}{\hbar} \int_{t_0}^t \mathrm{d}t' H_I(t', t_0)\ + \left (-\frac{i}{\hbar} \right )^2 \int_{t_0}^t \mathrm{d}t'' \int_{t_0}^t \mathrm{d}t''' H_I(t'', t_0)\,H_I(t''', t_0) +\dots $$
Going back to $U = e^{-iH_0t}U_Ie^{iH_0t}$, idenfitying $G_0 = e^{iH_0t}$ and $H_I = V$, we get the Dyson's equation as per the point above.
Hence $U$ is also called the Dyson's operator and the perturbative expansion known as Dyson series/expansion.
The time ordering operator $T$ should also be added.
Quantum field theory & example 4
While the first quantisation propagator is the same as a Green's function, in second quantisation the propagator is only termed as such by analogy.
Indeed: $$G^+(x,y) = \theta(x^0 - y^0) \langle \Omega|  \hat{\phi}(x)\hat{\phi}^\dagger (y) |\Omega \rangle, $$ 
where $|\Omega\rangle$ is the interacting vacuum, $x$ and $y$ are now four-vectors in spacetime, and $\hat{\phi^\dagger}(y)$ is a field operator creating a particle at $(y^0, \mathbf{y})$.
Your expression in just obtained by requiring the overlap $A$ between an incoming and outcoming state to be expressed in terms of the free, non-interacting states: $$ A = _{\mathrm{interacting}}\langle q|p\rangle_{\mathrm{interacting}} = _{\mathrm{free}}\langle q|S|p\rangle_{\mathrm{free}}, $$ with $S$ being the scattering matrix.
The scattering matrix $S$ is related to the evolution operator $U$ above by $S = \lim_{t\rightarrow \infty, t_0 \rightarrow -\infty} U(t,t_0)$.
