It is not an answer, but just some hints.
Consider the simplest free QFT with a massless bosonic scalar, the terms in the Lagrangian are local : $\phi(x) \square \phi(x)$. Considering an interacting theory ($\phi^3, \phi^4$). You are interested in calculate scattering amplitudes with incoming particles and outcoming particles. You will use the propagator in $\frac{1}{k^2}$ which form is direcly linked to the above Lagrangian term, and you may note that this propagator has a pole when $k$ is on-shell.
If you consider only a tree-level diagram, the transition amplitude is simply the product of propagators, each propagator could be written $\frac{1}{l^2}$, where $l$ is the sum of some external momenta (at each vertex, you have momentum conservation). So a pole of the scattering amplitude corresponds to the pole of the propagators, and this corresponds by putting on-shell some particular sum of the external momenta.
Now, consider a loop-diagram, with dimensional regularization, like $I(q) \sim g^2 (\mu^2)^\epsilon\int d^{4-\epsilon}p \frac{1}{p^2}\frac{1}{(p-q)^2}$,where $\epsilon$ is $>0$, $q$ is an external momentum. By using the Feynmann formula $\frac{1}{ab} = \int_0^1 \frac{dz}{[az+b(1-z)]^2}$, you will get : $I(q) \sim g^2 (\mu^2)^\epsilon\int_0^1 dz \int (d^{4-\epsilon}p) \dfrac{1}{[p^2 - 2p.q(1-z)+q^2(1-z)]^2}$, and finally :
$I(q) \sim g^2 (\mu^2)^\epsilon ~\Gamma(\frac{\epsilon}{2})\int_0^1 dz \dfrac{1}{[q^2 z(1-z) ]^{\large \frac{\epsilon}{2}}}$
Here, $\epsilon$ is $>0$, so we see that if $q^2=0$, the integral is not defined, so $q^2=0$ should represent a pole for the scattering amplitude.
The relation with the locality could be seen as looking at the Fourier transform (taking $\epsilon=0$) of the scattering amplitude which could be written $I(x) \sim [D(x)]^2$, where $D(x)$ is the propagator in space-time coordinates.
Now, we should hope that any scattering amplitude, with loops, should have poles, which corresponds to some particular sum of the external momenta being on-shell.
