Source Transformation and Power Output
You are right, source transformation doesn't preserve power dissipated by the source. 
Consider the following schematic. The circuits below are thevenin/norton equivalents of each other. Yet only one is dissipating power. What you will notice is that if you attach a load the load dissapates the same power in both circuits. 







simulate this circuit &ndash; Schematic created using CircuitLab

Let's say we attach an 300Ω load. 

The power dissipated by the load on the voltage source circuit can be found easily with P = V^2/R. 

The power dissipated by the load on the current source circuit can be found easily with P = I^2*R.



Plugging in (I skipped a voltage/ current division step), we get:

P = 0.75^2 / 300
and 
P = 0.0025^2 * 300 

which both evaluate to 1.875mW. 



Notice that the voltage and current source dissapate different amounts of power, just like in your example.

I encourage you to find the power dissipated in R13 and R14 in your example. You should find them equal. Also try a special case where your load resistance is the same value as the thevenin/norton resistance. Are the current/voltage source power dissipations still different?