LED forward voltages vary from 1 point something to 5V or even more. It depends for one on the color of the LED (which is determined by dopant material, so it's obvious why forward voltage will differ). So not all LEDs will behave the same when loaded with 3V. Some might not even light up. But most diodes do have less than 3V forward voltage, and at 3V they are going to be very bright and their life will be shortened.
And indeed what you were measuring is still the battery voltage minues the product of internal resistance and current. Real batteries are modelled as an ideal voltage source with zero resistance, in series with a (hopefully small) resistor which is its internal resistance. Voltage of ideal source equals that of non-loaded battery - as measured with high impedance multimeter. Equation for this circuit says that this voltage equals sum of internal resistance of the battery and the LED's resistance, times the current through the loop. If you measure current as well, you'll be able to find out internal resistance of battery as well as resistance of the diode AT THAT VOLTAGE. As thomas said, it is a non-linear device (you should be able to find I versus V plot for it in its datasheet) so resistance will vary depending on voltage applied. See here for examle:
A sample red LED datasheet
The steepness of the linear part of the curve (Fig. 2) determines the resistance of the diode in its normal operating region and further. You can estimate resistance as for example
(1.8V - 1.6V) / (30 - 7 mA) = 8.7 Ohm. You can see two things - first, if you keep increasing voltage, current will go up VERY fast. A diode might handle excessive current for a while depending on how bad it is but we all know what will happen in the end. Also note that as the current rises, so does the voltage drop on the internal resistance, therefore the actual voltage applied on the diode is less than one might think - 0.1V less in your case.
Without measuring current or knowing one of the parameters, you don't have enough information to determine resistances.
This also explains why a LED can be used as relatively stable voltage reference. If you put it in series with a resistor, and apply a voltage that is not stable, the current through the diode and resistor will vary, however, the voltage on the diode will vary much less than that. Steeper the curve, less the variation. In fact, let's calculate.
Take the operating point of LED to be say 1.7V. Current through diode is 20mA. Say the "unregulated" voltage is 10V. To get 20mA, we need 500Ohm resistance. Since diode resistance is 8.7Ohm, we need to add 491.3Ohm resistor in series with diode.
Say now that the voltage increases to 10.5V. Current through the LED/resistor will be 10.5V / 500Ohm (diode resistance is constant in this region), which yields 21mA. Extra 1mA over 8.7Ohm resistance yields only 8.7mV rise in voltage accross the diode. And so from 500mV change we got only 8.7mV change. In fact the rate of change of diode voltage will be equal to the proportion of diode resistance to the sum of diode and current setting resistor resistance. So 8.7 / 500 in this case. For any variation of input voltage, the diode voltage will change by that voltage multiplied with this ratio - 1.74%.