Okay, first we have the phenomenon: Yes. adding salt increases the boiling point of water, which means that you have to input more energy to get the water to boil, but your egg or pasta will cook faster once you do, because the water will be hotter.
Then there's the why. The boiling point of a liquid is the temperature at which the vapor pressure of the liquid is the same as the atmospheric pressure above the liquid. If we can artificially increase the vapor pressure of the liquid, we decrease the boiling temperature. If we can artificially decrease the vapor pressure of the liquid, we increase the boiling temperature. So the question has now become: why does the vapor pressure of water decrease when we add salt to it?
So imagine a pot of water. At any given temperature there will be some water molecules in the gas phase above the pot (that's the origin of the vapor pressure), and some in the liquid phase in the pot. The proportion in the two phases is determined by the interplay of lowering potential energy (by decreasing elevation in gravity, by forming hydrogen bonds, by lining up the polar ends of the molecules, etc.) and increasing the entropy (there's more accessible states in the gas phase, most liquids are incompressible, etc.). The potential energy part favors the liquid phase, while the entropy part favors the gas phase. The real requirement here is to minimize the free energy, F = U - TS, with F the free energy, U the potential, T the temperature, and S the entropy. Since S is paired with the temperature, increasing the temperature increases the impact of the entropy part, which is why the vapor pressure increases as we increase the temperature.
So now we toss in some salt, while keeping the temperature fixed. The volume fraction of the water decreases, and suddenly there are new accessible states for the water molecules in the liquid phase -- so the vapor pressure decreases. We keep adding salt and the vapor pressure keeps decreasing. If we keep going, eventually there's no vapor pressure.
Raoult's law says that the vapor pressure of a solution is proportional to the vapor pressure of the pure solvent (basically that there is a straight line between the pure vapor pressure and zero, when we've buried it in salt). That's taken as the definition of an ideal solution. Real solutions have a curved functional form between the two boundary conditions, with the deviations from linearity coming from interactions between the solute (the salt) and the solvent (the water). Those interactions might be things like breaking up the network of hydrogen bonds in the water, disrupting the polarization arrangement (both of which will favor gas phase), or bonding/pairing up with water molecules (which will favor liquid phase). At relatively low concentrations of solute the interaction effects are pretty small, so the dependence of vapor pressure on solute concentration remains roughly linear. The cool observation though is that at most temperatures and for most solvents, it doesn't matter what solute you use (as long as the solute itself doesn't have a vapor pressure), the vapor pressure of the solvent is still decreased by adding solute (which indicates that the entropic contribution is the most important part, and the interactions don't play a big role).
Now to sum up: for a given concentration of salt dissolved in water, there are more states accessible to the water molecules in the liquid phase than there are in pure water. So at every water temperature as we pour in energy to make it boil, there will be a lower vapor pressure than there would have been without the salt, and thus we won't get to the boiling point until the water has reached a higher temperature (until we've poured in more energy than we would have had to). Salt does disrupt the network of hydrogen bonds in the water molecules, but the effect isn't very big at reasonable concentrations of salt, and it's never big enough to counteract the entropic effect.
To figure out why this happens, you need to think about what boiling is, and how it works.
As you would know, the water in the pot boils because its temperature was raised above the boiling point by the flame. This required a net transfer of heat from the flame, through the pot, to the water in the pot. Why did the heat flow in this direction? Because the flame is hotter than the water in the pot, even when the water starts boiling ($T_{flame} > T_{boil}$)
Now, think about the water in the bottle. The only way for it to get heat is through the water in the pot. As long as the temperature of the water in the pot, $T_{pot}$, is less than $T_{boil}$, it is still liquid, and it transfers some heat to the water in the bottle. The water in the pot boils off at $T_{boil}$, and can no longer transfer heat as efficiently to the water in the bottle.
This effectively means that the water in the bottle is restricted to a maximum temperature of slightly less than $T_{boil}$, and that is why it never boils.
Another way to think of this is, there must be a temperature difference for a heat transfer to take place. Since the maximum possible temperature of the pot water is $T_{boil}$, the temperature of the bottle water can never exceed this.
EDIT: Another factor to consider is the low conductivity of glass, which means a high temperature difference is required to let a small heat flux through.
Best Answer
The following variables apply to this problem:
$Q$ - the amount of heat transferred into each liter of water
$m_{water}$ - the mass of water in each container
$m_{salt}$ - the mass of salt in each container
$C_p$ - the specific heat of the given mixture
$\Delta T$ - the temperature change of each liter of water
$t$ - time
For these variables, the following heat equation applies:
$\frac{Q}{t} = \frac{(m_{water}+m_{salt})C_p \Delta T}{t}$
This equation can be solved for time required to reach the boiling point, to give
$t = \frac{(m_{water}+m_{salt})C_p \Delta T}{Q/t}$
This equation can be set up for each container, and the ratio of the times it takes to reach the boiling point can be found by dividing one equation by the other, which conveniently eliminates some of the variables. At a constant heating rate for each container, that ratio becomes the following:
$\frac{t_1}{t_2} = \frac{(m_{water}+m_{salt1})C_{p1} \Delta T_1}{(m_{water}+m_{salt2})C_{p2}\Delta T_2}$
This means that the mass of salt in each container, the specific heat of each solution, and the temperature rise of each solution, must be known for the specific mixtures being tested before this question can be answered.