Your problem is not specific to <PERSON>'s PDE, but can be traced back to ODE's, even the simple harmonic oscillator. Say you want to solve: $$ \frac{d^2G}{dt^2}+G = \delta $$ Going to Fourier transform: $$ \tilde G(\omega)=\int\frac{d\omega}{2\pi}e^{i\omega t}G(t) \ G(t)=\int dte^{-i\omega t}\tilde G(t) \ (-\omega^2+1)\tilde G = 1 $$ You want to divide by $1-\omega^2$. However, when $\omega=\pm1$ this is zero, so the "value" there is undefined. Using the theory of distributions, you can prove that $\hat G$ is necessarily: $$ \tilde G = PV\frac{1}{1-\omega^2}+C_+\delta(\omega-1)+C_-\delta(\omega+1) $$ with $PV$ denoting the Cauchy principal value and $C_\pm$ being constants. The converse is easy to check, using the fact that: $$ f(t)\delta(t) = f(0)\delta(t) $$ Going back to time: $$ G = \frac{\sin |t|}{2} +C_+e^{-it}+C_-e^{it} $$ with $H$ you see that $C_\pm$ is specified by the initial conditions $G(t_0),\frac{dG}{dt}(t_0)$.
Back to your case, the similar situation happens when you divide by $k^2-\omega^2$. From: $$ (-\omega^2+k^2)\tilde G = 1 $$ you deduce: $$ \tilde G = PV\frac{1}{-\omega^2+k^2}+C_+(k)\delta(\omega-|k|)+C_-(k)\delta(\omega+|k|) $$ with now $C_\pm$ are functions defined on $R^3$. They can be determined by the Fourier transforms of $G,\partial_tG$ at $t_0$ which will give the unique solution to the Cauchy problem.
In physics, you are often interested in the case of causal solutions, i.e. $G$ is supported on positive times. It turns out, using <PERSON>'s lemma and the residue theorem, that it amounts to taking the limit of $\epsilon\to 0^+$ of: $$ \tilde G = \frac{1}{k^2-(\omega-i\epsilon)^2} $$ and more generally making the substitution $\omega\to \omega-i\epsilon$
Hope this helps.
Answer to comment
a) Actually the prescription retarded/advanced/Feynman will determine the $C_\pm$. 526
Once again, there is an ambiguity in defining $\frac{1}{k^2-\omega^2}$ which these prescriptions help resolve. If you want to interpret it as the Cauchy $PV$, then you need to add the appropriate $C_\pm$.
However, in this case, it is easier to think directly in terms of counter integration since you are not interested in solving the Cauchy problem. Indeed, writing explicitly what the prescriptions mean for the $C_\pm$ is needlessly complicated.
b) It depends. The problem arises when you divide by zero, in which case you need additional information to uniquely define your Gren's function (boundary condition, causality ...). However, this is not always necessary. Take for example the operator $-\Delta+m^2$ with $m$ real in arbitrary dimensions. Then its Green's function is (in Fourier): $$ \tilde G = \frac{1}{k^2+m^2} $$ which makes sense no matter what since the denominator never vanishes. Adding boundary conditions would overdetermine it.
An important caveat is that when you are taking Fourier transforms, you are restricting your attention to tempered solutions. In general, you could have more solutions thus requiring further restrictions. 976