There is a difference between the Schwinger effect and an effective action.
The Schwinger effect states that the QED vacuum is ''unstable''. If you apply a strong enough electrical field you would be able to rip apart electron positron pairs which are randomly generated everywhere.
An effective action on the other hand is usually a subset of a full QFT. For example Chiral perturbation theory is an effective theory for QCD.
The Schwinger effective action is the exact action in the sense that it takes into account all quantum effects including perturbative and nonperturbative ones. It can be thought of as the quantum version of classical action. You can derive the equation of motion for the quantum field theory based on the Schwinger effective action.
I did not know that it was called Schwinger effective action, but just effective action. At least it is called that way in Peskin Schroeder.
Anyhow those equations Chun Yang stated are called Dyson-Schwinger or Schwinger-Dyson equations and in principle they are an exact solution for a given QFT.
However concerning pair production you should look into the Schwinger effect.
The Schwinger effect is the pair production from vacuum in the presence of external field. This is a non-perturbative effect and was first proposed in the context of QED. But it should appear in many cases, including the period of inflation in cosmology, quark production in heavy ion collision. In the former example, the external field is gravity; in the latter case, strong gluon fields induce the quark pair creation.
If the effective action is real, the vacuum-defined by the path integral over the fluctuating, not the external, fields-is stable, since it changes by a phase, upon introducing the fluctuating fields. If the effective action has an imaginary part, the vacuum is unstable and the imaginary part defines the decay rate per unit time and volume.
In a ``strong enough'' external and constant electric field the effective action for QED does acquire such an imaginary part and this can be described as the instability of the vacuum of zero particle density to the vacuum of finite particle density-the external field creates pairs at a finite rate, given by the imaginary part. This is known as the Schwinger effect. In a constant electric field it's possible to perform the path integral over the fermions exactly, because it's possible to solve the Dirac equation exactly-and, since the fermions enter quadratically, obtain an exact result. It's not that pair production is non-perturbative, it's that the calculation can be done without recourse to perturbation theory, in this case. In other cases, where the Dirac equation can't be solved exactly, it is, of course, possible to perform a perturbative calculation, that's valid when perturbation theory is.
Resummation in qcd is non-perturbative or perturbative? It's not exact! Semiclassical instanton approach is non-pertubative or perturbative? It's also not exact!