If you measure the current with the motor stalled you'll probably get more like 2.5A, at least until the motor begins to smoke or your supply current-limits.
The simplest useful model of a DC motor is a resistance (the one you measured) in series with a voltage source that bucks the applied voltage as the motor spins. That "back EMF" is proportional to how fast the motor is spinning. If you were to spin the motor with another motor, you'd measure that voltage at the open motor terminals (it would act as a generator). The resistance represents the losses in the copper windings mostly.
If the motor was perfect, no friction in the bearings or air resistance on the spinning armature, the current with no load on the shaft would go down to zero as the motor winds up from a start. As it turns out, for your motor, it goes from 2.5A to 0.15A so about 94% of the way there. As you load the motor, the armature RPM drops and the current increases.
If your motor runs at (say) 5,000 RPM with no load, and 5V applied, there is 150mA so the back EMF must 5V-300mV = 4.7V. So back EMF = (RPM/5000)* 4.7V.
If you load the shaft so that the RPM drops to 3,000 RPM, the back EMF will be 2.82V and therefore the current will be
I = (5V-2.82V)/2\$\Omega\$ = 1.09A.
Here is a speed/torque/efficiency curve for a random brushed DC motor (Mabuchi RS-555SH):
As you can see, as the speed drops (under load, with fixed voltage applied) the current increases. The \$\eta\$ (efficiency) peaks at relatively low current, but the maximum power output will be close to where \$\eta\$ = 50%.
If you start applying fast-varying voltage (as in microseconds or milliseconds) to the motor you also have to consider the inductance, but for slow-varying 'DC' it doesn't matter.