Simulating matching op-amp input impedance with offsets
With the output connected directly to IN- there is about a 30mV
  difference between the input and the output, where the input is about
  400mV.


Measuring a high impedance node with a voltmeter can result in a significant error. The input node impedance is around 783 kohm and your meter might have an input impedance of around 10 Mohms so, when you put your meter at the input your new effective impedance will be: -

$$\dfrac{1}{\frac{1}{783k} + \frac{1}{10M}}$$

This equals 726 kohm and would naturally reduce 400 mV to 371 mV. Here's your 30 mV difference.


  However to get the circuit to provide a gain of 1 I doubled the
  feedback resistor to 1.5Meg. The result is much better and the output
  error is much smaller. Why is it that double the resistor was needed
  here and not 750K?


Fix the basic problem of not using your meter in parallel with a large impedance. I would also choose a much better op-amp than the LM324 because it has an input offset voltage of 3 mV and if you require accuracy, 3 mV may be too much of an extra uncertainty. Also, if you require good AC performance you might find the LM324 somewhat unable to deliver the goods. Also there is no mention in the LM324's data sheet what resistive input impedance it has over and above the leakage currents from either input pin.


  is there a better way to model this?


Model the effective input resistance of your meter at the op-amp +INPUT.