You have exact equations for the solution in the related question Time it takes for temperature change. Here I would add a few comments.
It is actually easier if container is thick! Then suppose that all water is at same temperature $T$ and all the air in the frizer is at the same temperature $T_e$. $T_e$ is constant.
If that is so, you can use only Fourier's law to describe how heat $Q$ leaves the container
$$\frac{\text{d}Q}{\text{d}t} = \frac{\lambda A}{d} (T-T_e).$$
$d$ is thickness, $A$ area and $\lambda$ thermal conductivity of the container.
Knowing that water cools as heat is leaving the container
$$\text{d}Q = m c \text{d}T,$$
where $m$ is mass and $c$ is specific heat capacity of the liquid, you get rather simple differential equation
$$m c \frac{\text{d}T}{\text{d}t} = \frac{\lambda A}{d} (T-T_e),$$
$$\frac{\text{d}T}{(T-T_e)} = \frac{\lambda A}{d m c } \text{d}t = K \text{d}t,$$
which has exponential solution:
$$K t = \ln \left(\frac{T-T_e}{T_0-T_e}\right),$$
$$T = T_e + (T_0-T_e) e^{-Kt}.$$
Adding salt to water makes it freeze at a lower temperature. This fact is being used in two different ways in the two scenarios you mention. Dissolving sodium chloride in water is slighly endothermic, but this effect is small and to the best of my knowledge isn't important in the drink cooling process.
Putting salt on the highway is quite straightforward: we don't want ice to form, so we put salt in the water to prevent that. This doesn't just change the amount of time it takes ice to form, it actually completely prevents ice from forming, unless the temperature gets so low that the water can freeze even with salt in it.
Cooling your drink is a bit more complicated, because in this case the rate at which things happen is important. You don't want your drink to be less than $0^\circ C$ because it would freeze; instead you want to cool it down to a few degrees Celsius nice and quickly.
The rate at which it cools depends on two things: the temperature of its surroundings (the colder the better) and the heat conductivity between it and them. You could try to cool it by putting it in a bowl of ice at $0^\circ C$, but the problem is that the ice is solid and will only touch the bottle at a few points. This results in a poor thermal conductivity, so the drink will only cool slowly.
To get around this, you could try mixing the ice with some water. Now the bottle is touching the liquid over a large surface area, and the liquid itself has a higher thermal conductivity than solid ice due to mixing, so heat will be transferred much more quickly. But the problem is that the water won't be at zero degrees any more, at least not at first (I'm assuming the water comes from a tap, so it's not chilled initially). You have to wait for quite a bit of the ice to melt before the water's temperature will drop. Also, once you put your warm drink into the water it will heat the water up as the drink cools down, so again you have to wait for the ice to melt in order for the water to cool again.
The solution to this is to make the ice melt faster. You can do this by adding salt. This lowers the freezing point, making the water less "happy" about being in the liquid state, so it melts more quickly. This means firstly that the tap water you've added to the ice will cool to close to $0^\circ C$ much more quickly, and secondly that once you've put your drink in there the water will stay cold as the ice continues to melt.
It's also possible that, with the salt added, the water can go to below $0^\circ C$, but this will only happen if the ice is quite a bit colder than $0^\circ C$. This could be the case, but my intuition is that the rate of cooling due to the ice melting faster is more important here than the final temperature. You could easily test this by putting a thermometer in the salty ice water and seeing if it goes much below freezing.
There's also the fact that dissolving the salt is endothermic, as you mentioned. To test whether this is important, you could try adding salt to some chilled water without any ice, and see if the temperature drops a lot. My feeling is that it will only drop by a tiny amount that will be hard to measure with a normal kitchen thermometer, but you can always try the experiment.
Best Answer
You probably had a hard time finding an answer because it isn't straightforward. There are many factors which can influence the best approach, and the times.
For example, the temperature of the surroundings and the insulation of your container will have significant effects on the temperature over time. A highly insulated container will allow you to reach a lower minimum temperature, and cooler surroundings will do the same. Placing the bottle in the freezer will also lower the minimum possible temperature (even if you take it out of the freezer to do this experiment, a cold bottle is better than a warm one).
The initial temperature of the drink compared to the initial temperature of the ice are also very important.
To determine when it will be the coldest, you need to compare the amount of heat transferred from the ice to the drink, and from the surroundings to the drink. As long as the rate of heat transferred to the ice is greater than the rate of heat transferred from the surroundings to the drink, the system will be cooling. I would expect to find the lowest temperature near the point when the ice is melting. This is because melting ice requires the latent heat of fusion, and ice retains it's temperature during this process; so the ice can absorb heat while still remaining at the same temperature. This phase transition represents a fairly large portion of the cooling ability of ice.
The specifics of when this will happen can vary a lot depending on your setup though. Factors such as the geometry of the container and the geometry of the ice can be significant in determining the heat transfer between those objects and their surroundings (for example, surface area effects the rate of heat transfer, and geometry can effect convection currents).