If we consider no Faradaic reactions happening on either electrode then all you will have is a capacitor.
Imagine you have some mobile ions in solution, then these ions will feel an electric field and start to migrate. There will be an opposite driving force of diffusion against the concentration gradient that has built up (on the negatively polarized electrode positive ions will accumulate and vice versa on the positively polarized electrode). This is what we call an electrical double layer, some charge on the electrode which is compensated by mobile ions in the electrolyte. At some point there will be a balance between the driving force of diffusion, the charge separation creating an electric field countering the electric field you applied, and the applied electric field leading to no net movement of the ions. The total electric field will not be constant between the electrodes as near the electrodes significant charge separation takes place which will strongly counteract the applied electric field. Making the electric field in between the electrodes outside the double-layer region near 0. The characteristic time to reach this situation is the same as for a normal capacitor.
One thing to consider is how strong all of these effects are, charge separation at a large distance generates an incredibly strong electric field: Poisson's equation: $\nabla\cdot \vec{E}=\frac{\rho}{\epsilon}$, with $\vec{E}$ the electric field, $\rho$ the charge density in C/m³ (to convert from moles to C use Faraday's constant 96485 C/mol), and $\epsilon$ the permittivity of the medium. Try plugging in µM concentration differences over cms and you will get very high voltages. So when you say the ion concentration in the bulk will be 0, it is very unlikely unless your concentration is very small and the distance between the electrodes is very short.
If you would like to know if ions are remaining in between the electrodes it depends on what accuracy you need. You can simply sample some liquid and measure the conductivity, but depending on your application it might not suffice.
Finally, if you have no ions at all then you will simply have a regular capacitor with a constant electric field between the electrodes.
I hope this answers your question.