Impedances to simplify RC circuit?
You are mixing Laplace transforms (\$s\$-variable) with phasors (\$j\omega\$-variable). Please see: Under what conditions does jw equal the laplace variable s in an electrical circuit?. Both are called impedances, but you either use one or the other at the same time.

I share what I would consider the easy way of solving that circuit. If we call \$Z_1\$ to the parallel impedance of \$R\$ and \$C_1\$, and \$Z_2\$ to the impedance of \$C_2\$, we get a simple voltage divider with the following transfer function:

$$H(s) =\frac{V_o}{V_i}= \frac{Z_2}{Z_1 + Z_2}$$

The impedances are:
$$Z_1 = R \parallel (C_1s)^{-1} = \frac{R \times \frac{1}{C_1s}}{R + \frac{1}{C_1s}} = \frac{R}{1 + RC_1s}$$
$$Z_2 = (C_2s)^{-1}$$

Resulting in:
$$H(s) = \frac{\frac{1}{C_2s}}{\frac{R}{1 + RC_1s} + \frac{1}{C_2s}}$$

As you see, this equation is identical to the last one you wrote, but with \$s = j\omega\$ \$(I(s)\$ can be cancelled\$)\$.

Simplifying we get:
$$\boxed{H(s) = \frac{RC_1s + 1}{R(C_1 + C_2)s + 1}}$$
and, depending on the ROC of the transfer function:
$$\boxed{H(j\omega) = \frac{RC_1j\omega + 1}{R(C_1 + C_2)j\omega + 1}}$$



Edit: All this is valid if the circuit is initially relaxed. In this case if the capacitors are initially discharged. See s-domain equivalent circuits and impedances.