Would this Solid State Relay work to control my portable electric heater?
That seems to be a good choice. 

Note that at 1500/120= 12.5 A and a drop voltage of 1.6V (at 25C !!) the SSR will dissipate 20 Watt. Here is a datasheet with some more details. The temperature/current graph indicates that a case temperature of 100C should be OK. Assuming a maximum ambient temperature of 50C (or can you guarantee less?) that means you need a heatsink of 50 / 20 = 2.5C/W. If find it a bit troublesome that there is no indication of the drop voltage at let's day 150C.

A last note: I hope for you that your ebay source sells the genuine product not some counterfeit.

(edit: heatsink calculations)

Temperature and heat (=power) behave like voltage and current:

   Temperature (degees C)    =&gt; Voltage (V)
   Heat (power, W)           =&gt; Current (A)
   Thermal resistance (C/W)  =&gt; Electrical Resistance (Ohm)


In your case the 'voltage' is 50C, the 'current' is 20W, hence the 'resistance' is 50/20=2.5C/W. That is the maximum thermal resistance (between case and ambient air) that will keep the temperature difference at or below 50C. 

A good datasheet will give you the RthCase-Ambient without any heatsink, so you can calculate what you can dissipate without heatsink. I did not find that in the SSR datasheet, maybe because it was never intended to be used without heatsink. So, you need a heatsink this a maximum Rth of 2.5C/W. For the calculation I ignored the Rth between the case and the heatsink. For a TO220 I recall figures ~0.5C/W, I assume it is much lower for your SSR.

Note that the calculation assumes that the ambient air temperature does not rise above 50C. YOU must make that true. No amount of heatsinking will do any good in an airtight enclosure.