BJT circuit analysis confusion
I assume there is a DC path for emitter current that isn't shown and, for some reason, doesn't significantly change the AC circuit.

The first equation is correct.  Looking out of the base, there is Thevenin signal source with Thevenin voltage

$$V_{tb} = V_{sig}\frac{R_B}{R_B + R_{sig}}$$

and Thevenin reistance

$$R_{tb} = R_B||R_{sig}$$

The signal base current is then given by

$$I_b = \frac{V_{tb}}{R_{tb} + \left( \beta + 1\right)\left(r_e + \frac{1}{sC_E} \right)}  = V_{sig}\frac{R_B}{R_B + R_{sig}}\frac{1}{R_B||R_{sig} + \left( \beta + 1\right)\left(r_e + \frac{1}{sC_E} \right) }$$

Your approach should give the same equation after some algebra.