The generalized delta is just an antisymmetrizer. You just need to use the commutation identity that you wrote at the end of the first displayed equation to slide the $A$'s through the antisymmetrizer.
Edit with a few more details:
I said the generalized delta
$$
\delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}=\sum_{\sigma\in S_k} {\rm sgn}(\sigma)
\delta^{i_{\sigma(1)}}_{j_1}\cdots \delta^{i_{\sigma(k)}}_{j_k}
$$
is an antisymmetrizer because if you contract it to a general tensor $T_{i_1,\ldots,i_k}$
and this way define a new tensor
$$
W_{i_1,\ldots,i_k}:= \delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}\ T_{j_1,\ldots,j_k}
$$
(I used Einstein's convention, but I don't care about upstairs/downstairs)
what you get is an antisymmetric tensor, i.e., $W_{i_1,\ldots,i_k}$ changes sign if any two indices are switched.
Now the commutation, or sliding of $A$'s through the antisymmetrizer is the identity
$$
\delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}A^{j_1}_{h_1}\cdots A^{j_k}_{h_k}=
A^{i_1}_{j_1}\cdots A^{i_k}_{j_k}\delta^{j_1,\ldots, j_k}_{h_1,\ldots, h_k}\ .
$$
As for the proof of the wanted identity, one can start as in the OP (a bad move that needs fixing)
$$
\frac{1}{k!} A^{i_1,\ldots,i_k}_{h_1,\ldots,h_k} (A^{-1})^{h_1,\ldots,h_k}_{j_1,\ldots,j_k}=
\frac{1}{k!} \left(
A^{i_1}_{p_1}\cdots A^{i_1}_{p_k}\delta^{p_1,\ldots, p_k}_{h_1,\ldots, h_k}\right)
\left(\delta^{h_1,\ldots, h_k}_{q_1,\ldots, q_k}(A^{-1})^{q_1}_{j_1}\cdots (A^{-1})^{q_k}_{j_k}\right)
$$
but the $A$'s and $A^{-1}$'s are on the wrong side of their attached generalized delta. So use the commutation twice to rewrite this as
$$
\frac{1}{k!}
\left(\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k}A^{p_1}_{h_1}\cdots A^{p_k}_{h_k}\right)
\left((A^{-1})^{h_1}_{q_1}\cdots (A^{-1})^{h_k}_{q_k}\delta^{q_1,\ldots, q_k}_{j_1,\ldots, j_k}\right)
$$
$$
=\frac{1}{k!}\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k}
(A^{p_1}_{h_1}(A^{-1})^{h_1}_{q_1})\cdots
(A^{p_k}_{h_k}(A^{-1})^{h_k}_{q_k})
\delta^{q_1,\ldots, q_k}_{j_1,\ldots, j_k}
$$
$$
=\frac{1}{k!}\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k}
\delta^{p_1}_{q_1}\cdots\delta^{p_k}_{q_k}
\delta^{q_1,\ldots, q_k}_{j_1,\ldots, j_k}
=\frac{1}{k!}\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k}
\delta^{p_1,\ldots, p_k}_{j_1,\ldots, j_k}
=\delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}\ .
$$
Edit with even more details:
For completeness, let me prove the identity about the last two contracted generalized deltas giving a single one.
By definition, and summing over permutation $\sigma,\tau$ in the symmetric group $S_k$,
$$
\delta^{i_1,\ldots,i_k}_{p_1,\ldots,p_k}\delta^{p_1,\ldots,p_k}_{j_1,\ldots,j_k}=
\sum_{\sigma,\tau}{\rm sgn}(\sigma){\rm sgn}(\tau)
\delta^{i_{\sigma(1)}}_{p_1}\cdots\delta^{i_{\sigma(k)}}_{p_k}
\ \delta^{p_{\tau(1)}}_{j_1}\cdots\delta^{p_{\tau(k)}}_{j_k}
$$
But, by permuting factors,
$$
\delta^{p_{\tau(1)}}_{j_1}\cdots\delta^{p_{\tau(k)}}_{j_k}=
\delta^{p_1}_{j_{\tau^{-1}(1)}}\cdots\delta^{p_k}_{j_{\tau^{-1}(k)}}\ .
$$
We insert this equality and do the summation over the $p$ indices and get
$$
\delta^{i_1,\ldots,i_k}_{p_1,\ldots,p_k}\delta^{p_1,\ldots,p_k}_{j_1,\ldots,j_k}=
\sum_{\sigma,\tau}{\rm sgn}(\sigma){\rm sgn}(\tau)
\delta^{i_{\sigma(1)}}_{j_{\tau^{-1}(1)}}\cdots\delta^{i_{\sigma(k)}}_{j_{\tau^{-1}(k)}}
$$
$$
=\sum_{\sigma,\tau}{\rm sgn}(\sigma){\rm sgn}(\tau)
\delta^{i_{\sigma(\tau(1))}}_{j_1}\cdots\delta^{i_{\sigma(\tau(k))}}_{j_k}
$$
by reordering the factors again. To finish, use
${\rm sgn}(\sigma){\rm sgn}(\tau)={\rm sgn}(\sigma\tau)$ and notice that each permutation $\rho=\sigma\tau$ appears exactly $k!$ times.
Best Answer
First of all we put $$ \Delta^{i_1,...,i_k}_{j_1,...,j_k}:=\text{det}\begin{pmatrix}\delta^{i_1}_{j_1}&\cdots&\delta^{i_1}_{j_k}\\ \vdots&\ddots&\vdots\\ \delta^{i_k}_{j_k}&\cdots&\delta^{i_k}_{j_k}\end{pmatrix} $$ for each choice of indices so that we have to prove that $$ \delta^{i_1,...,i_k}_{j_1,...,j_k}=\Delta^{i_1,...,i_k}_{j_1,...,j_k} $$ for each choice of indices.
So if $\delta^{i_1,...,i_k}_{j_1,...,j_k}$ is zero then either one between the $k$-tuples $(i_1,...,i_k)$ and $(j_1,...,j_k)$ has duplicate indices or the indices of such $k$-tuples are not equal so that in the first case the matrix $\Delta^{i_1,...,i_k}_{j_1,...,j_k}$ has two rows duplicate and in the second case the same matrix has at least one column zero and thus the equality tritivally holds by notable properties of the determinant. Conversely if $\delta^{i_1,...,i_k}_{j_1,...,j_k}$ is not zero then the indices of the $k$-tuples are distinct and the indices of the first $k$-tuple are the same of the second so that the first $k$-tuple is a permutation of the second and vice versa. So we know that $$ \Delta^{i_1,...,i_k}_{j_1,...,j_k}=\sum_{\sigma\in\mathfrak S_k}\text{sgn}(\sigma)\delta^{i_1}_{j_{\sigma(1)}}\delta^{i_2}_{j_{\sigma(2)}}...\delta^{i_k}_{j_{\sigma(k)}} $$ and so we analyse which and how many terms of the previous sum are not zero. So by the zero-product property we know that $\text{sgn}(\tau)\delta^{i_1}_{j_{\tau(1)}}\delta^{i_2}_{j_{\tau(2)}}...\delta^{i_k}_{j_{\tau(k)}}\neq0$ for some $\tau\in\mathfrak S_k$ if and only if the quantity $\delta^{i_h}_{j_{\tau(h)}}$ is not zero and this happens if and only if $i_h=j_{\tau(h)}$ for each $h=1,...,k$ so that the only term not zero in the previous summ is that related to the perutation $\tau$ and thus $$ \Delta^{i_1,...,i_k}_{j_1,...,j_k}=\sum_{\sigma\in\mathfrak S_k}\text{sgn}(\sigma)\delta^{i_1}_{j_{\sigma(1)}}\delta^{i_2}_{j_{\sigma(2)}}...\delta^{i_k}_{j_{\sigma(k)}}=\\\text{sgn}(\tau)\delta^{i_1}_{j_{\tau(1)}}\delta^{i_2}_{j_{\tau(2)}}...\delta^{i_k}_{j_{\tau(k)}}=\text{sgn}{\tau}=\delta^{i_1,...,i_k}_{j_1,...,j_k} $$ as we desired.