Putting 2 loads in parallel, will not lower the current.
In this case the current will be doubled (because they are equal loads), resulting in 15 mA.
But i guess TS is trying to predict the current at 5 volts compared to the given current at 12 volts, and is using the formula U=IR.
But that is not what that formula is about.
If any, given U and I, try P=UI.
12*0.0075= 0.09 Watts.
I=P/U; 0.09/5 = 0.018 A
So 18 mA.
But still, if you are going to measure, you will find out that this is not true.
A LED needs a current, and when that is met it will light up driving the SSR.
But there is also an internal protection sketched as a resistor so all bets are off on this.
Who knows what actually is in there.