(This is too long to fit into a comment.)
Note that the product in the integrand of the right side foils out as
$\sum_{A\subseteq [n]} (-1)^{n-|A|}\left(\prod_{i\not\in A} (i-1)\right)\prod_{j\in A}K(x_j,x_j).$
(Here $[n]=\{1,2,\dots,n\}$.) On the other hand, we can use Leibniz formula to expand the determinant in the left hand side to obtain
$\iint\cdots\int \sum_{\sigma\in S_n} (-1)^{\sigma}\left(\prod_{i=1}^nK(x_i,x_{\sigma(i)}) \right)dx_1\cdots dx_n$ Interchange the order of summation and integration. Observe that $K(x_1,x_{\sigma(1)})\cdots K(x_n,x_{\sigma(n)})$ can be reordered into a product of factors of the form $K(x_a,x_b)K(x_b,x_c)\cdots K(x_d,x_a)$ (this follows from the cycle decomposition of the permutation $\sigma$), and the integration of this over $x_b,x_c,\dots,x_d$ is, by induction on the two properties at the bottom of the OP, smply $K(x_a,x_a)$. From here we can once again switch the order of summation and integration.
I think the last ingredient needed is something from the representation theory of the symmetric group. In other words, we need to know that the number of ways a permutation $\sigma$ can be decomposed into cycles with representatives $i_1,i_2,\dots\in A\subseteq [n]$, weighed by sign, is the coefficient of $\prod_{j\in A}K(x_j,x_j)$ in the integrand's polynomial on the right hand side (or something roughly to this effect).