Complex Exponential Narrow Band Channel We can describe a BPSK signal, x, as:

$$x=cos(2\pi f_c t+n\pi),\;n\in \{ 0,1 \}$$

The BPSK signal is then transmitted through a narrow band frequency flat Rayleigh channel, described by a single complex exponential $h=Ae^{i\theta}$. $A$ has Rayleigh distribution and $\theta\sim\mathrm{Unif}(0,2\pi)$. The received signal at the destination is given as:

$$y=hx=Ae^{i\theta}[cos(2\pi f_c t+n\pi)].$$

My question is, what does the exponential term of the channel $e^{i\theta}$ represent? Clearly:

$$y\neq A[cos(2\pi f_c t+n\pi + \theta)].$$ This can be verified by expanding the sinusoids terms into their complex exponential forms.
The equations are a bit wrong. For a signal $x=cos(2\pi f_{c}t+n\pi), n\in \left\{0,1\right\}$ transmitted through a complex exponential channel $h=Ae^{i\theta}$, the received signal $y$ is not $hx$. $h$ is in the complex baseband notation and your signal $x$ is in the bandpass notation. 

To answer your actual question first, $\theta$ is the phase change that is incurred by the signal when it passes through the channel i.e. the delay. And the actual received signal is $y=Re\left\{x_{l}Ae^{i\theta}e^{i2\pi f_ct}\right\}$, where $x_{l}$ is the baseband representation of our signal $x$ which in our case would be something like $x_{l} = e^{i2\pi\frac{(n-1)}{2}}$, where $n\in \left\{0,1\right\}$.