Under what circumstances would the large internal input resistance of an actual op amp affect circuit operation?
Each opamp parameter which normally is neglected during calculation will affect the operation (the gain) of the amplifier (input and output impedances, finite and frequency-dependent open-loop gain,...). 

However, in most cases we do not care about these effects because they will be either not too important (because the error is acceptable) or are overshadowed by external tolerances of the feedback network and/or  parasitic influences caused by the hardware realisation (pin capacitances,...).  

This is a typical example for the fact that in electronics no formnula is correct by 100%. It is simply not possible - and it makes no sense - to include all possible known physical effects in our expressions, functions and formulas. And it is one of the most challenging engineering tasks to decide if - for a specific application - the "simplified" expression may be applied with sufficient accuracy or not. 

As far as the finite input impedance is concerned, we try to follow a general rule which requires that each of the external resistors should be small if compared with this input resistance. However, what means "small"? Factor 10 or 100 or 1000 ? The answer simply depends on the required accuracy of the gain value. But in most cases, the resistor tolerances are more important.   

As an example: Here is the gain expression (non-inverting) for resistive feedback (ideal: 1+R2/R1) if input and output resistances are taken into account.

G=N/D  with 

N=(Eo R2 Rin + Eo R1 Rin + R1 Rout)

D=(Eo R1 Rin + Rin Rout + R2 Rin + R1 Rout + R1 Rin + R1 R2)

Note that the open-loop gain Eo is set to a fixed value. You can imagine how the expression would look like in case of a realistice frequency-dependent gain expression for a two pole model:

Eo(w)=Eoo/[(1+s/w1)(1+s/w2)]