If you use more or bigger batteries, then you can increase the battery life.
The important thing here is the energy of the batteries. Your device uses 5W of power, which means that it uses 5Wh of energy every hour of operation.
If you use a battery that has 10Wh of energy capacity, then in theory, you device would work for two hours (in practice a bit less due to various inefficiencies).
A DC-DC converter can help here, that is, it can help you use more batteries in series to have more total energy. You can also connect the device (using a Dc-DC converter) to a car battery (~840Wh) and have it work for ~168 hours.
The DC-DC converter changes the voltage and current, but laves power constant, so, it can take 12V/!A and produce 1V/12A. In practice, due to inefficiencies, the converter will always take more power than it outputs, the difference being used to heat up the converter itself.
However, you have to use a switching DC-DC converter, and not a linear regulator such as a 7805. The linear regulator takes as much current from the power source as the load needs (it would use 12V 1A (12W) to produce 5V 1A (5W) dissipating the rest (7W)), therefore it dissipates much more power as heat.
A switching converter is usually 70% to 95% efficient.
Car batteries have a high cold cranking current, as Neil says in his answer.
But they are usually a poor choice for solar storage, where deep cycle or leisure batteries are preferred - they are better able to withstand deep discharge.
If leisure batteries have a CCA (cold cranking amps) rating it is likely to be lower. They should have a max current rating instead. Or, their capacity may be stated at different discharge rates such as C/10 (which would discharge them in 10 hours). C (1 hour) and 10C (6 minutes, which may only deliver half the capacity at C/10).
C is their nominal capacity, which is often measured at a discharge rate of C/20, i.e. 5A for a 100 Ah battery.
Or the max discharge rate may be expressed as a C rate (10C) instead of a current.
Multiply this rate by the capacity to get the current.
So if you have a battery which allows 10C discharge, and you need 78A, then its capacity must be at least 7.8 Ah. And preferably several times that if you want a reasonably long battery life.
Best Answer
The fan will draw less current when it is given less volts. The current draw is roughly proportional to applied volts as is the speed. Consult the fan manufacturers specs and/or check on a variable lab supply to obtain a minimum voltage for reliable operation.
Lower fan voltages will mean that less air is shifted. Now make some PWM to control the fan and you will save battery power. If you run the fan at say 70% speed meaning about 70% voltage and about 70% current then power is halved if your proposed PWM system is perfect.
What you can do is connect your PWM control to a temp sensor so the fan speeds up when hot shifting its full specified amount of air and slows down when cold saving the battery and making less noise and increasing bearing life. Sure this is more complicated than a resistor but these days it is worth it when SMD is used.