Comprehension problems with op-amps
First, a classic op amp amplifies the DIFFERENCE in the voltages at the + and - terminals.
First and a half, an op amp has a VERY high voltage gain.
Second, every linear circuit that uses an op amp provides a feedback path from the output to the - input.
This creates a control loop, where the whole point of the feedback path is to drive the difference between the two inputs to zero. In normal operation, the op amp in circuit actively drives that difference to zero. For analysis of the circuit, then, it is generally good enough to assume that the difference is zero.
Start with the classic noninverting buffer stage, where the input is fed to the + input and the output is tied to the - input. The only output value that will yield zero difference is Vout = V+ (the voltage at the + input).
Consider the classic unity gain inverting buffer, where the + input is tied to ground, the signal is applied through a resistor to the - input, and an equal resistor ties the - input to the output. The op amp will try to drive the - input to ground. If the input signal is above ground, current must flow in the input resistor, and attempt to flow into the - input. This would induce an offset voltage across the op amp input impedance, which is amplified by the op amp, and that gives rise to an output voltage, which in turn gives rise to current through the feedback resistor. If the current in the input resistor is equal to the current in the feedback resistor, then there is no current left to induce an offset voltage at the - input, and the amp is happy.
Typical opamps have DC gains in the 10^5 to 10^6 region. This means if the output is within the rails, say 0 to 5v, or +/- 15v, then the input will indeed be measured in microvolts.
As you say, this is sufficiently close to zero to be deemed to be zero for many purposes.
One of the purposes for which input=0v is a good enough approximation is when solving for the DC gain of a fed back amplifier. Typically, several resistors will draw current from input and output voltages measured in volts, and sum their currents at one of the amplifier input terminals. Whether that terminal has a voltage of 0uV, or 10uV, is irrelevant for most purposes, as the error is parts per million.
Typically, amplifier input offset errors will be measured in mV, so for accurate systems, we have to worry about input offsets long before we have to worry about whether the input voltage is really zero or not.
For an ideal opamp, where the gain is infinite, then the input voltage is zero, theoretically.
All calculations are not based on zero differential input voltage, only approximate calculations. It's usually a pretty good approximation at DC if the op-amp is a precision type (high gain, low input offset voltage, and low input bias currents and offset current relative to resistor values). Often it is less than the input offset voltage (usually not with 'zero-drift' types, however).
However, if you want to account for finite open-loop op-amp gain (and for offset voltage and CMRR) you need to make a bit more complex calculation, and in any case, you have to keep in mind that the differential input voltage is only approximately zero.
Since open-loop gain drops with frequency (typically at -20dB/decade above something like 10Hz), it's not hard to have a significant closed-loop gain error at higher frequencies, especially if you have a higher frequency and/or higher ideal closed-loop gain.