Answer by periblepsis for Finding transfer function of high-pass circuit

Well, first off \$R_{_\text{S}}+\frac1{s\,C_1}\ne\frac{s \,R_{_\text{S}}\,C_1}{s \,R_{_\text{S}}\,C_1+1}\$. Rather, \$R_{_\text{S}}+\frac1{s\,C_1}=\frac{s \,R_{_\text{S}}\,C_1+1}{s \,R_{_\text{S}}\,C_1}\$.

Next, the transfer function for this high-pass must follow this template:

$$\frac{v_o}{v_i}=A\cdot\frac{s}{s+\omega_{_0}}$$

In your case, \$A=\frac1{1+\frac{R_{_\text{S}}}{R_1\,\mid\mid \,R_2}}\$ and \$\omega_{_0}=\frac1{C_1\,\cdot\,\left[R_{_\text{S}}+\left(R_1\,\mid\mid \,R_2\right)\right]}\$.

You should be able to see both of these. As \$\omega\to\infty\$, the capacitor becomes a wire and you just have a voltage divider. So the value of \$A\$ should be obvious from that fact. Separately, swap the series order of \$R_{_\text{S}}\$ and \$C_1\$ and you can see that \$C_1\$sees a load of \$R_{_\text{S}}+\left(R_1\,\mid\mid \,R_2\right)\$. So \$\omega_{_0}\$ is similarly obvious.

The zero is at \$s=0\$ and the pole is at \$s=-\omega_{_0}\$.

Let's start with your first equation, which you correctly formed:

$$\begin{align*}\frac{v_o}{v_i}&=\frac{R_1\mid\mid R_2}{R_{_\text{S}}+\frac1{s\,C_1}+\left(R_1\,\mid\mid \,R_2\right)}\\\\&=\frac{s\,C_1\cdot\left(R_1\,\mid\mid \,R_2\right)}{1+s\,C_1\cdot\left[R_{_\text{S}}+\left(R_1\,\mid\mid \,R_2\right)\right]}\\\\&=\frac{C_1\cdot\left(R_1\,\mid\mid \,R_2\right)}{C_1\cdot\left[R_{_\text{S}}+\left(R_1\,\mid\mid \,R_2\right)\right]}\cdot\frac{s}{s+\frac1{C_1\cdot\left[R_{_\text{S}}+\left(R_1\,\mid\mid \,R_2\right)\right]}}\\\\&=\frac{R_1\,\mid\mid \,R_2}{R_{_\text{S}}+\left(R_1\,\mid\mid \,R_2\right)}\cdot\frac{s}{s+\frac1{C_1\cdot\left[R_{_\text{S}}+\left(R_1\,\mid\mid \,R_2\right)\right]}}\\\\&=\frac{1}{1+\frac{R_{_\text{S}}}{R_1\,\mid\mid \,R_2}}\cdot\frac{s}{s+\frac1{C_1\cdot\left[R_{_\text{S}}+\left(R_1\,\mid\mid \,R_2\right)\right]}}\\\\&=A\cdot\frac{s}{s+\omega_{_0}}\end{align*}$$

So we come full circle back to where I started, above.

So \$f_{0}=\frac{\omega_{_0}}{2\pi}\approx 111.038\:\text{Hz}\$ and \$A\approx 0.93\approx -628\:\text{mdB}\$.

An LTspice run finds:

I couldn't get an exact \$45^\circ\$ out of it. But I bracketed it. It shows that \$111.0133\:\text{Hz}\lt f_{0}\lt 111.04525\:\text{Hz}\$.

The value of \$A\$ can be worked out by adding \$3.0103\:\text{dB}\$. Here, I would get something like \$-0.628\:\text{dB}\$. Which is about what I said, earlier.