Why is Ie = Ic + Ib but IR2 is not = IR1 + Ib?

During transistor dc conduction Ie splits unequally at the emitter-base junction with the majority of emitter's current going to collector and then it's what's left of Ie that makes up the transistor's total dc base current Thus Ie must = Ic + Ib.

IR1(dc) consist of the dc current through R2 joined by the transistor's dc base current at the R2-base junction,

Thus IR1 dc must = (IR2 + IB)dc

Consequently;

IR2 dc = (IR1 - Ib)dc

Ie is Ic+Ib and IR2 is IR1 - Ib with the ripple produced by Vin riding on it.

I don't know how you determined what you said but remember that if you measured it with a meter there is a current burden to any measurement which may be mis-leading. Look up the impedance of your voltmeter in the range you are using to make the measurement.

IR2 IS equal to IR1 + ib, with due regard to polarity.

actually Ib = IR1 – IR2, which means

IR2 = IR1 – Ib

By Kirchhoff's Law, the sum of the currents flowing in and out of any junction is 0. I gather that IR2 and IR1 are currents flowing one way or the other (you didn't say which) through R1 and R2. There's also a current flowing through C1. The total of all four currents at the junction are 0, not just three.