If what happens when the input is open is important, then you can follow the answer from WhatRoughBeast (also see comment below). If not, you can simplify things a bit:
Get rid of D1, R1, R2
Change R3 to 20K and connect to the +5V reference voltage
Change R5 to 20K
That will have an output of 0V with an input of -2.5V and +5V with an input of 2.5V.
The difference is the the output voltage with the input disconnected will be about 7.5V vs. 5.0V with WRB's circuit.
If that's important, you can still simplify it a bit:
Get rid of D1, replace R3 with a short and increase R1, R2 to 20K each. Leave R5 and R6 at 10K. This will be equivalent (0V out for -2.5V in, 5V out for 2.5V in, 5V out for input open), but will draw less current from the reference input (and have one fewer component).
Most op-amps should have a spec for supply current in the datasheet. This current flows from the positive supply pin to the negative supply pin, so it loads both of your supplies.
In addition you need to consider the current flowing from the op-amp's output pin into the load and the feedback network. If the op-amp is sourcing current, the output pin current comes from the positive supply, and loads only that supply. If the op-amp is sinking current, the output pin current flows to the negative supply, and loads only that supply.
You need to determine the output pin current by considering the operating behavior of the circuit, the load impedance, and the current that will be drawn by the feedback network.
Best Answer
No, a normal op-amp cannot output voltages that are not between the supply rails. Some can get very close (millivolts) and others can only get within volts of the rails.