The mistake depends by what you are trying to achieve.
In this case, you have the voltage source that is suppose to simulate the forward-biased diode. This means that, since the source is connected with the plus to ground, you are generating -0.7V at the non-inverting input of the OPAMP. So, a current is flowing from ground across the source, and that current depends on the output voltage and the value of the top resistance (perhaps 1 Ohm).
Then, let's look at the inverting input and the voltage divider. Since between the two OPAMP inputs there is a virtual short circuit, the central point of the divider will be -0.7V. Using two equal (1 Ohm I guess) resistors, you are causing the output to be at -1.4V. Again, the current will flow out of the ground.
Now, back again to our generator. We said that the non-inverting input is at -0.7, and the output is at -1.4V. Hence we will have a 0.7V drop over the resistor, and since (as guessed before) it's a 1 Ohm resistor, the current across the resistor and the generator/diode (since the OPAMP inputs are ideally open circuits) will be 0.7A
Conclusion
If you are trying to simulate the 0.7V drop of the forward-biased diode, it's what the supply is doing. If you are expecting to see positive voltages, it's not because of negative resistors, but because the supply has to be flipped.
Update
There are two cases, depending on the initial state:
The output of the OPAMP is HIGH: then, the diode is reverse biased, no current is flowing in the upper branch, and the non-inverting input is at a higher voltage than the inverting, that is always at half the output voltage. Hence, the OPAMP goes into positive saturation;
The output is LOW: then, the diode is forward biased, the voltage at the non-inverting pin is -0.7, and the situation is the aforementioned.
You have 500 mV headroom, so a LDO should be possible. A diode is not a exact voltage drop. The output of a LDO will be much better regulated. Unless this is a very high volume product and the extra few cents actually matters, use a LDO.
Best Answer
Yes, if you raise the LDOs GND pin, the output will rise accordingly. This is because the LDO now see's a higher voltage between it's reference point and ground.
All a fixed regulator is, is a standard regulator with the programming resistive divider inside the IC, rather than outside it.
If you look at the above image (imagine the resistive divider is inside the IC), you can see if you raise the bottom of the Vref pin, the output voltage will have to rise in order to match the divider input of the comparator with the divider input. As Photon comments though, it's not so accurate though to use a diode (unless you use a more complex circuit which can compensate for temp variations, etc, but then you may as well buy an adjustable regulator)
In the LM7805 datasheet, there is an example of how to create and adjustable regulator by driving the reference input with an opamp:
And a slightly simpler version which uses a fixed divider and no opamp:
For this version, you can use the formula provided - if you use a 5V regulator and want 5.3V, then:
5V * (1 + (3kΩ / 50kΩ)) = 5.3V
So R2, = 3kΩ, and R1 = 50kΩ (if you want exact you need to take into account the error from the Iq*R2 bias current, which is also given in the formula - the above is just a basic "near enough" example)