From the figures you give, for the gauge input shorted to ground, by my calculation the gauge has an internal resistance of 12/0.09 = ~130 Ohms connected between the input and 12V. That will certainly skew the expected output from the opamp and make it unusable as is.
So, for 'full' the gauge draws 90mA. For 'empty' it will draw (12-5.3)/130 = ~ 45mA.
What you need is a circuit which translates 0V to 45mA and 5V to 90mA. That could be simply a transistor and a few resistors. I'll have a think and come up with something!
So, for 'full' the gauge draws 90mA. For 'empty' it will draw (12-5.3)/130 = ~ 45mA.
What you need is a circuit which translates 0V to 45mA and 5V to 90mA. That could be simply a transistor and a few resistors. I'll have a think and come up with something!