pi model of common collector
To be honest I do not understand your problem. It seems that you are overthinking the problem. Stick to one single model and use it for all configurations (CC, CE, CB).
For example, you can use T-model. Thus for CC (emitter follower) amplifier, it will look like this:

In this model \$r_e\$ is equal to:
$$r_e = \frac{V_T}{I_E} = \frac{\alpha}{g_m} = \frac{r_{\pi}}{\beta +1}$$
And we alredy see that the voltage gain of a voltage follower is:
$$\frac{V_{OUT}}{V_{IN}} = \frac{R_E}{r_e + R_E}$$
We can use this model also for CE amplifier

For this circuit we have
$$V_{OUT} = -I_CR_C$$
$$V_{IN} = I_E\:r_e + I_E\:R_E$$
Aditional we know thet \$I_C = I_B*β\$ and \$I_e = I_B + I_C = I_B + I_B\:β = I_B(β + 1)\$
therefore \$ \large \frac{I_C}{I_E} = \frac{I_B\:β}{I_B(β + 1)} = \frac{β}{β + 1}\$
From this, we can write that \$I_C = I_E\frac{β}{β + 1}\$ thus we have:
$$V_{OUT} = -I_CR_C = -I_E\:R_C \:\frac{β}{β + 1}$$
And the voltage gain is:
$$\frac{V_{OUT}}{V_{IN}} = \frac{-I_E\:R_C \:\frac{β}{β + 1}}{I_E\:r_e + I_E\:R_E} = -\frac{R_C}{r_e +R_E} \:\frac{β}{β + 1}$$
As you can see we can use the same small-signal model for all amplifier configurations.
Of course, we can use a voltage-controlled current source model as well.

For example, the input resistance of this circuit is:
$$R_{IN} = \frac{r_e + R_E}{1 - g_m\:r_e} = (\beta +1)(r_e + R_E)$$
As homework try to prove that this formula is true.
Also, we can use a hybrid-pi model as well, see this example of CC amplifier
KVL equations for this small signal model