Why do we use an opamp to produce a voltage output above 1V in this circuit?

It's all down to speed. What your circuit doesn't show is the self-capacitance of the photodiode: -

Given that the signal produced by the photodiode is current (Iph shown above), if this is rapidly changing like in an optical data receiver, the junction capacitance will have a significant effect on rise and fall times.

However, with a transimpedance amplifier we are, in effect, shorting out the capacitance and now, the current signal takes the path of lowest resistance and that is into the virtual earth node of the inverting input. This vastly improves high frequency performance.