From a thermodynamics point of view, I'd say you should leave the water in.
Temperature is a measure of the active kinetic energy of the molecules in a substance. Warming up is essentially the surrounding environment imparting some of its kinetic energy into the object being warmed up. Simply thinking about that, the more you have that needs warming, the more energy it requires to warm, and so the slower its temperature will rise (given the same rate of exchange of thermal energy - the water submerges at least some of the other contents, so if anything this is an overestimate). Now consider that water has a relatively high specific heat, meaning that it takes more energy to warm it up. The food inside the cooler won't be warmed up faster than the surrounding water, so since it takes more energy to heat the food and water than just the food, so the food will stay cooler longer.
The Igloo (yes, the cooler company) FAQ supports this view:
During use, it is not necessary to drain the cold water from recently melted ice unless it is causing contents to become soggy. The chilled water, combined with ice, more readily surrounds canned and bottled items and will often help keep contents colder more effectively than the remaining ice alone.
Don’t drain cold water – Water from just-melted ice keeps contents cold almost as well as ice and preserves the remaining ice much better than air space. Drain the water only when necessary for convenient removal of cooler contents or before adding more ice.
I think the key point a lot of people forget is that what matters isn't how long the ice lasts, but how long the contents remain under some temperature.
Assuming convection is fairly significant relative to the rate of energy input from the outside (a good assumption, I think), it doesn't matter how good an insulator air is, the inside temperature will be the same throughout, necessitating that the rate of heat energy coming into the cooler is the same in both cases, so the ice will melt at the same rate, and once the ice is gone, the cooler with water will take much longer to warm. I'm still working on the no-convection theory (which would provide at best an extreme overestimate), but in the meantime if anyone wishes to posit that convection is tiny enough to bridge the enormous gap (assuming I find the upper bound does eclipse water), please explain why you believe so.
Some math/physics to back this up for the quantitatively inclined. (This would be so much easier with the MathML markup from the math and physics sites.)
The cooler will be very near perfect convection, the heat is entering the cooler slowly enough that the contents - air, water, and ice - are at the same temperature (namely 32°F/0°C/273.15K). Heat conduction,
H, as far as our coolers are concerned depends only on
H = kAΔT/x, where
k is the thermal conductivity of the cooler,
A is the area of the cooler through which thermal energy is flowing,
x is the thickness of the cooler, and
ΔT is the temperature difference between the inside and outside of the cooler (
T_out - T_in). Notice that all these are the same for both coolers. Now, melting the ice requires energy of
Q = Lm where
Q is the total energy required,
L is the (latent/specific) heat of fusion, and
m is the mass of ice in the cooler. Since the total energy is
Q = Ht, we can calculate the time required to melt all the ice:
Q = Ht = kAtΔT/x -> t = Qx/(kAΔT) = Lmx/kAΔT. Since all variables are the same for both coolers, it will take the exact same amount of time for the ice to melt.*
*: We are ignoring the air replacing the ice, which would actually give a (very) slight advantage to retaining water. Draining the water in that cooler requires adding heat - the excess heat of the air that replaces the melted water. Fortunately, that excess heat is fairly easy to calculate:
m = Vρ -> V = m_ice/ρ_ice = m_air/ρ_air -> m_air = m_ice * ρ_air/ρ_ice. The air comes in with an excess energy of
Q = m_air*C_air*ΔT = m_ice*(ρ_air/ρ_ice)*C_air*ΔT. This reduces the energy required from the cooler heat influx, reducing the time required:
Q = Lm_ice = Q_cooler + Q_air = kAtΔT/x + m_ice*(ρ_air/ρ_ice)*C_air*ΔT -> t = (L/ΔT - ρ_air/ρ_ice)*C_air)mx/kA. The fractional difference this will make ends up being about
4e-6*ΔT, or about
0.016% on a rather hot (40°C) day, coming to 2min 18s over 10 days. So we were right to ignore it.