The crux of the proof is to indeed consider the Schur power matrix. Let $A$ be an $n\times n$ matrix. Then consider the $n! \times n!$ matrix $S(A)$ indexed by permutations $\sigma, \tau$ such that the $(\sigma,\tau)$-entry is given by $\prod_{i=1}^n a_{\sigma(i),\tau(i)}$ for $\sigma, \tau \in \mathfrak{S}_n$.

Schur's theorem. Let $A$ be a positive semidefinite matrix. Then, for any vector $x$, $x^TS(A)x \ge \det(A)x^Tx$.

It turns out that actually the determinant of $A$ is also an eigenvalue of $S(A)$ (and from the above theorem, it is the smallest eigenvalue). From this the Immanantal inequality also follows as a corollary. Let me repeat a proof of this claim below (this is not my original proof, obviously).

Proof. Consider the tensor product matrix $\otimes^n A$, and augment $x$ (which is a vector of dim $n!$) by padding with zeros appropriately to obtain a vector $z$ that is of dim $n^n$, so that $$x^TS(A)x= z^T(\otimes^n A)z.~~~~~~~~~~~~~~~(*)$$ Since $A$ is psd, we can write $A=B^TB$ for a lower-triangular matrix $B$. Thus, we have (using simple properties of the tensor product) $$[z^T(\otimes^n A)z = z^T(\otimes^n B^TB)z = z^T(\otimes^n B^T)(\otimes^n B)z.$$

Now by Cauchy-Schwarz we obtain \begin{align*} x^TS(A)x & \le \left[z^T(\otimes^n B^T)^2z\right]^{1/2}\left[z^T(\otimes^n B)^2z\right]^{1/2}\\ &= \left[z^T(\otimes^n (B^T)^2)z\right]^{1/2}\left[z^T(\otimes^n B^2)z\right]^{1/2}. \end{align*} Using the relation $(*)$ twice on the rhs above we thus obtain \begin{equation*} x^TS(A)x \le [x^TS(B^2)^Tx]^{1/2}[x^TS(B^2)x]^{1/2}. \end{equation*} Since $B^2$ is lower triangular, the diagonal entries of $S(B^2)$ are all easily seen to be actually equal to $\det(B^2)=\det(B)^2$. Moreover, $S(B^2)$ itself is lower-triangular, so that $x^TS(B^2)x = \det(B)^2x^Tx$ which equals $\det(A)x^Tx$. Substituting this in the final inequality above, the proof is complete.

Many thanks to Denis for pointing out my erroneous initial "proof". This time around the proof is correct, and directly proves the assertion in line 3 of the OP, i.e., $\chi(e)\det(A)\le d_\chi(A)$ (I will write $d_\chi(I)$ instead of $\chi(e)$ for uniformity).

The explicit notation is cumbersome, so I am just writing a proof sketch.

1. First, recall that $d_\chi(A)=z^T(\otimes^n A)z$ for a suitable vector $z$
2. Next, use Cauchy-Schwarz to obtain $$|z^T(\otimes^n (X^TY))z|^2 = |z^T(\otimes^n X^T)(\otimes^n Y)z|^2\le z^T(\otimes^n X^TX)z \cdot z^T(\otimes^n Y^TY)z$$
3. Now write $A=C^TC$ for some upper triangular matrix $C$ (since $A$ is PSD we can do this). Then, put $X=C$ and $Y=I$ above, to obtain
4. $|z^T(\otimes^n C)z|^2 = |z^T(\otimes^n I)z|^2|\det C|^2 \le |z^T(\otimes^n C^TC)z|\cdot |z^T(\otimes^n I)z|$, where we used the upper triangular nature of $C$ for the first step. In other words, we have shown that
5. $d_\chi(I)^2 \det(A) \le d_\chi(A)d(I)$, since $|\det C|^2=\det(C^TC)=\det(A)$.