Understanding spectra in polyphase filter banks
Since you define the input to the filter $p_1[n]$ as $x_1[n]=x[2n-1]$ and the channel impulse response as $p_1[n] = p[2n+1]$ the spectrum at the output of that channel will be as follows:

First express $X_1(e^{j\omega})$, the DTFT of the input to the channel no 1.

$$ x[n] \longrightarrow \boxed{z^{-1}} \longrightarrow v[n]=x[n-1] \longrightarrow \boxed{\downarrow2} \longrightarrow w[n]=x[2n-1] $$

Now as can be seen, $V(e^{j\omega}) = X(e^{j\omega}) e^{-j\omega} $ and also $W(e^{j\omega}) = \frac{1}{2} \sum_{k=0}^{1} V( e^{j( \frac{\omega-2\pi k}{2} )  })$ When combined it yields;

$$ X_1(e^{j\omega}) = \frac{1}{2} \sum_{k=0}^{1} X( e^{j( \frac{\omega-2\pi k}{2} )  }) e^{-j( \frac{\omega-2\pi k}{2} )} $$ 

In this case

$$ X_1(e^{j\omega}) = \frac{1}{2} e^{-j\frac{\omega}{2}} \cdot(X(e^{j\frac{\omega}{2}}) - X(e^{j(\frac{\omega}{2} - \pi)})) $$ 

Note: This is where the linked paper an therefore image four in the question is confusing. It does not include the linear phase part $e^{-j\frac{\omega}{2}}$ in equation (7). Without this linear phase part, as mentioned in the question, the signal is not a valid factor two down sampled signal. The phase term 'fixes' the equation to make the phase 'repeat' properly according to the new sample frequency.

That's the signal spectrum, and the filter spectrum similarly is:

$$ h[n] \longrightarrow \boxed{z^{1}} \longrightarrow v[n]=h[n+1] \longrightarrow \boxed{\downarrow2} \longrightarrow w[n]=h[2n+1] $$

$$ H_1(e^{j\omega}) = \frac{1}{2} \sum_{k=0}^{1} H( e^{j( \frac{\omega-2\pi k}{2} )  }) e^{j( \frac{\omega-2\pi k}{2} )} $$

In this case $$ H_1(e^{j\omega}) = \frac{1}{2} e^{j\frac{\omega}{2}} \cdot(H(e^{j\frac{\omega}{2}}) - H(e^{j(\frac{\omega}{2} - \pi)})) $$ 

Note: The linear phase term has opposite sign to the phase term in $X_1$ so these terms will cancel each other when we multiply $X_1$ and $H_1$.

Hence the spectrum at $B_2$ is:

$$B_2(e^{j\omega}) = H_1(e^{j\omega}) X_1(e^{j\omega}) $$
$$ B_2(e^{j\omega}) = \left( \frac{1}{2} \sum_{k=0}^{1} H( e^{j( \frac{\omega-2\pi k}{2} )  }) e^{j( \frac{\omega-2\pi k}{2} )} \right)  \left( \frac{1}{2} \sum_{k=0}^{1} X( e^{j( \frac{\omega-2\pi k}{2} )  }) e^{-j( \frac{\omega-2\pi k}{2} )} \right) $$

in this case:

$$ B_2(e^{j\omega}) = \frac{1}{4} \cdot (X(e^{j\frac{\omega}{2}})H(e^{j\frac{\omega}{2}}) - X(e^{j\frac{\omega}{2}})H(e^{j(\frac{\omega}{2} - \pi)}) -X(e^{j(\frac{\omega}{2} - \pi)})H(e^{j\frac{\omega}{2}}) + X(e^{j(\frac{\omega}{2} - \pi)})H(e^{j(\frac{\omega}{2} - \pi)})) $$ 

For $A_2$:

$$ A_2(e^{j\omega}) = \frac{1}{4} \cdot (X(e^{j\frac{\omega}{2}})H(e^{j\frac{\omega}{2}}) + X(e^{j\frac{\omega}{2}})H(e^{j(\frac{\omega}{2} - \pi)}) +X(e^{j(\frac{\omega}{2} - \pi)})H(e^{j\frac{\omega}{2}}) + X(e^{j(\frac{\omega}{2} - \pi)})H(e^{j(\frac{\omega}{2} - \pi)})) $$

Note that the aliasing cancellation happens at the reconstruction part of the system: As can now be seen $A_2 + B_2$ results in one channel, whereas $A_2 - B_2$ will produce the other channel.