Solving all-pass using superposition
I agree with the comments that most op amp-based all-pass filters show a pull-down resistor after the capacitor rather than having it connected to the output as in Leach's example. Nevertheless, we can try to determine the transfer function of this circuit applying superposition. The trick to split \$V_i\$ in two separate sources is good to solve these types of circuits. We thus have two circuits to which we can apply superposition also. The first circuit appears below:



What you need to do is determine the value of \$\epsilon\$ in all configurations: write the voltage at (-) and (+) nodes and write \$\epsilon=V(+)-V(-)\$. In the first configuration in which \$V_o\$ is considered as an individual source and set to 0 V while \$V_i\$ is on, you have \$\epsilon_1=-\frac{V_i}{2}\$ while \$\epsilon_2=V_o\frac{1}{1+sRC}-\frac{V_o}{2}\$. Sum up these two values and solve \$0=(\epsilon_1+\epsilon_2)\$ considering an infinite-gain op. If everything goes well, you have \$V_{01}(s)=V_i(s)\frac{1+sRC}{1-sRC}\$.

For the second part, ground the left side of \$R_i\$ and consider the below circuit:



If you follow the above steps, then you should obtain \$V_{02}(s)=V_i(s)\frac{2sRC}{sRC-1}\$. You can now write that \$V_o=V_{o1}+V_{o2}\$ and if you do the maths ok, then you find \$V_o=V_i\$ confirming the comments left by the two readers. Now, if you correctly connect the resistor after the cap. to ground, you have the following circuit:



You can split \$V_i\$ as recommended and you can inspect the circuit as the first configuration is an inverting configuration having a gain of -1 while the second one is a non-inverting configuration have a gain of \$1+\frac{R_1}{R_1}\$ and amplifying the differentiator output.



So the output is equal to \$V_o=V_i(-1+\frac{sRC}{1+sRC}(1+\frac{R_1}{R_1}))\$ which is what Mr. Leach wrote but from a wrong sketch unfortunately. If you solve this expression, you have \$H(s)=-\frac{1-\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}\$ in which \$\omega_z=\omega_p=\frac{1}{RC}\$.

This expression is made of a RHP zero and a LHP pole tuned at the same frequency canceling each other in magnitude. It gives a flat 0-dB response. However, as the RHP zero lags in phase, it sums up with that of the LHP pole and you create a delay. This is the practical realization of the 1st-order Padé approximant of \$H(s)=e^{-s\tau}\$ with the pole/zero pair tuned at \$\frac{2}{\tau}\$.