Let $g=h$ be in $C_1$. Then they are conjugate, so $\sum_i\chi_i(g)\overline{\chi_i(g)}=|C_G(g)|$. But that's $\sum_i|\chi_i(g)|^2=24$, and the sum is $1^1+1^2+2^2+3^2+x^2$. So that should get you $x$ (well, you also have to know that $\chi_i(g)$ is a positive integer for $g$ in $C_1$).
Now can you do the row orthogonality to get $y$?
The number of irreducible representations of $G = S_4$, which is equal to the number of conjugacy classes of $S_4$, is $n = 5$.
The conjugacy classes of $G$ have sizes
$$
w_1 = 1 \,, \quad
w_2 = 6 \,, \quad
w_3 = 8 \,, \quad
w_4 = 6 \,, \quad
w_5 = 3 \,.
$$
On the vector space $ℂ^n$ we can now consider three different inner products:
The standard inner product
$$
⟨x, y⟩
= \sum_{i = 1}^n \overline{x_i} y_i
= \overline{x_1} y_1 + \overline{x_2} y_2 + \overline{x_3} y_3 + \overline{x_4} y_4 + \overline{x_5} y_5 \,.
$$
The scaled inner product
$$
⟨x, y⟩_{\mathrm{scal}}
= \frac{1}{|G|} ⟨x, y⟩
= \frac{1}{|G|} \sum_{i = 1}^n \overline{x_i} y_i
= \frac{1}{24} ( \overline{x_1} y_1 + \overline{x_2} y_2 + \overline{x_3} y_3 + \overline{x_4} y_4 + \overline{x_5} y_5) \,.
$$
The “characteristic” inner product, which is not only scaled but also incorporates the numbers $w_i$ as weights:
$$
⟨x, y⟩_{\mathrm{char}}
= \frac{1}{|G|} \sum_{i = 1}^n w_i \overline{x_i} y_i
= \frac{1}{24} ( \overline{x_1} y_1 + 6 \overline{x_2} y_2 + 8 \overline{x_3} y_3 + 6 \overline{x_4} y_4 + 3 \overline{x_5} y_5 ) \,.
$$
The condition $A A^{\mathsf{t}} = 𝟙$ mean that the rows of $A$ are orthonormal with respect to the standard inner product $⟨-,-⟩$.
Similarly, your proposed condition $A A^{\mathsf{t}} = |G| 𝟙$ means that the rows of $A$ are orthonormal with respect to the scaled inner product $⟨-,-⟩_{\mathrm{scal}}$.
However, the orthonormality of irreducible characters tells us that the rows of $A$ are orthonormal with respect to $⟨-,-⟩_{\mathrm{char}}$ instead.
For example, the last two rows of $A$ are given by $x = (3, -1, 0, 1, -1)$ and $y = (2, 0, -1, 0, 2)$.
These rows are orthogonal with respect to $⟨-,-⟩_{\mathrm{char}}$ since
\begin{align*}
⟨x, y⟩_{\mathrm{char}}
&=
\frac{1}{24}
\Bigl(
1 ⋅ 3 ⋅ 2 + 6 ⋅ (-1) ⋅ 0 + 8 ⋅ 0 ⋅ (-1) + 6 ⋅ 1 ⋅ 0 + 3 ⋅ (-1) ⋅ 2
\Bigr) \\[0.5em]
&=
\frac{1}{24} ( 6 + 0 + 0 + 0 - 6 )
=
0 \,.
\end{align*}
Both $x$ and $y$ are normalized with respect to $⟨-,-⟩_{\mathrm{char}}$ because
$$
⟨x, x⟩_{\mathrm{char}}
=
\frac{1}{24}
\Bigl( 3^2 + 6 ⋅ (-1)^2 + 8 ⋅ 0^2 + 6 ⋅ 1^2 + 3 ⋅ (-1)^2 \Bigr)
= \frac{1}{24}( 9 + 6 + 0 + 6 + 3 )
= 1
$$
and also
$$
⟨y, y⟩_{\mathrm{char}}
=
\frac{1}{24}
\Bigl( 2^2 + 6 ⋅ 0^2 + 8 ⋅ (-1)^2 + 6 ⋅ 0^2 + 3 ⋅ 2^2 \Bigr)
= \frac{1}{24}( 4 + 0 + 8 + 0 + 12 )
= 1 \,.
$$
Best Answer
Here is an alternative proof:
Lemma: The characters of irreducible representations form a basis for the space $\mathbb C_{class}(G)$ of class functions $f:G\to \mathbb C$.
Proof: We prove that whenever a class function $f$ is orthogonal to every character of irreducible representations, then it is the zero map. Let $\rho:G\to\text{GL}(V)$ be an irreducible representation of $G$. If $f\in \mathbb C_{class}(G)$, then the map $\phi=\sum_{g\in G}f(g)\rho(g^{-1})$ is a representation homomorphism. By Schur's Lemma $\phi=\lambda I$ for some $\lambda\in \mathbb{C}$. The trace of $\phi$ vanishes by assumption. Thus $\phi$ is the zero map in any irreducible representation and therefore in any representation of $G$. Setting the representation $\rho $ as the regular representation we have $\sum_{g\in G}f(g)\rho(g^{-1})=0$, and all group actions in the regular representation are linearly independent. Then $f=0$. $\blacksquare$
I sketch what is needed to prove your result, you can fill in the details.