Putting power supplies in parallel probably won't work, but putting them in series may help.
So, if you put two 12 V supplies in series to get 24 volts, you would then allocate 25 % of the 24 volts (ie 6 volts) to a resistor and the rest to the LEDs.
Your LEDs have about 3.5 volts across them, so you could have 5 in a string to use 17.5 volts and then the remaining 6.5 volts for a resistor. (6.5 Volts at 500 mA would mean you need a 13 ohm resistor)
You will notice that this isn't even as many LEDs as you have now.
The problem is that if you put in more LEDs and have less voltage across the resistor, then small variations in the supply voltage have a large effect on the LED current.
For example, if you have 3 LEDs with 3.5 volts across them and a 12 volt supply, and a 2.7 ohm resistor, there will be (12 - 10.5 volts / 2.7 ohms) or 555 mA flowing.
If the supply voltage went up to 13 volts, there would be (13 - 10.5 volts / 2.7 ohms) or 925 mA flowing.
The LED voltage would rise slightly, but this effect can make the LED brightness vary a lot or even destroy LEDs.
It can be reduced by allocating more of the supply voltage to the resistor and less to the LEDs.
There are dedicated modules for driving high powered LED arrays like these.
For example, this one:
http://www.ebay.com.au/itm/10x-LED-Power-Supply-Module-Drivers-1x3w-580-600mA-/290591133691?pt=LH_DefaultDomain_0&hash=item43a894cffb
I can't vouch for the quality, but it appears that each driver would provide about 600 mA to one LED but at greater efficiency than using resistors to limit the current.
There are 10 drivers in this package for $10.
It uses 12 Volts AC to drive the modules.
Unfortunately, the efficiency is not given, but it may be something like 80%.
So, a 3.5 volt LED drawing 600 mA would be using 2.1 watts.
This would take 2.1 watts * 100 / 80 or 2.625 watts from a 12 volt supply or 218 mA.
So, you would gain a little in efficiency, but the current through the LEDs would be regulated and so the brightness would not be highly dependent on the supply voltage.