$\newcommand{\ep}{\varepsilon}$Let us write $n$ instead of $N$ and $k$ instead of $n$. For brevity, let us also write $X_i$ and $Y_i$ instead of $X_{n,i}$ and $Y_{n,i}$.
For any function $f$ of several variables, let
\begin{equation*}
E_n f(X,Y,\ldots):=\frac1n\sum_{i\in[n]}f(X_i,Y_i,\ldots),
\end{equation*}
where $[n]:=\{1,\dots,n\}$.
It is given that the $X_i$'s are permutation invariant, the $Y_i$'s are permutation invariant, the $Y_i$'s are independent of the $X_i$'s,
\begin{equation*}
E_n X^k\to a_k,\quad E_n Y^k\to b_k \tag{10}\label{10}
\end{equation*}
for $k=1,2,\dots$. Here and in what follows, the convergence is in probability as $n\to\infty$.
We have to show that then
\begin{equation*}
E_n XY\overset{\text{(?)}}\to a_1 b_1. \tag{20}\label{20}
\end{equation*}
Note that $E_n XY=E_n(X-a_1)Y+a_1E_n Y$ and $E_n(X-a_1)^k=\sum_{j=0}^k\binom kj (-a_1)^j E_nX^{k-j}$. So, replacing $X_i$ by $X_i-a_1$, we see that without loss of generality (wlog) $a_1=0$. Similarly, wlog $b_1=0$. So,
\begin{equation*}
a_1=b_1=0, \tag{25}\label{25}
\end{equation*}
and \eqref{20} becomes
\begin{equation*}
E_n XY\overset{\text{(?)}}\to 0. \tag{20a}\label{20a}
\end{equation*}
For each small enough real $\ep>0$, let
\begin{equation*}
M:=M_\ep:=\ep^{-2/3}. \tag{27}\label{27}
\end{equation*}
Let
\begin{equation*}
\hat X_i:=X_i\,1(|X_i|\le M),\quad \check X_i:=X_i-\hat X_i=X_i\,1(|X_i|>M),
\end{equation*}
\begin{equation*}
\hat Y_i:=Y_i\,1(|Y_i|\le M),\quad \check Y_i:=Y_i-\hat Y_i=Y_i\,1(|Y_i|>M).
\end{equation*}
Note that
\begin{equation*}
P(|E_n XY|>4\ep)\le p_1+\cdots+p_4, \tag{30}\label{30}
\end{equation*}
where
\begin{equation*}
p_1:=P(|E_n \hat X\hat Y|>\ep),\quad p_2:=P(|E_n \hat X\check Y|>\ep),
\end{equation*}
\begin{equation*}
p_3:=P(|E_n \check X\hat Y|>\ep),\quad p_4:=P(|E_n \check X\check Y|>\ep).
\end{equation*}
Next,
\begin{equation*}
|\check Y_i|\le Y_i^4/M^3, \quad |\check X_i|\le X_i^4/M^3, \tag{40}\label{40}
\end{equation*}
and hence
\begin{equation*}
p_2\le P(E_n|\check Y|>\ep/M)\le P(E_n Y^4>M^2\ep)\to0, \tag{50}\label{50}
\end{equation*}
by \eqref{10} with $k=4$. Similarly,
\begin{equation*}
p_3\to0. \tag{60}\label{60}
\end{equation*}
Next, in view of the Cauchy--Schwarz inequality and the inequalities
\begin{equation*}
\check X_i^2\le X_i^4/M^2, \quad \check Y_i^2\le Y_i^4/M^2, \tag{40a}\label{40a}
\end{equation*}
we get
\begin{equation*}
p_4\le P(E_n \check X^2\,E_n \check Y^2>\ep^2)
\le P(E_n \check X^2>\ep)+P(E_n \check Y^2>\ep) \\
\le P(E_n X^4>M^2\ep)+P(E_n Y^4>M^2\ep)
\to0. \tag{70}\label{70}
\end{equation*}
So, \eqref{20a} reduces to
\begin{equation*}
p_1\overset{\text{(?)}}\to0. \tag{20b}\label{20b}
\end{equation*}
Note now that
\begin{equation*}
P(|E_n\hat X|>2\ep)\le P(|E_n X|>\ep)+P(|E_n\check X|>\ep) \\
\le P(|E_n X|>\ep)+P(E_n X^4>M^3\ep)\to0,
\end{equation*}
by \eqref{10}, \eqref{25}, \eqref{40}, and \eqref{27}.
So, $E_n\hat X\to0$. Also, $|E_n\hat X|\le M$. So, by dominated convergence,
\begin{equation*}
E(E_n\hat X)^2\to0. \tag{80}\label{80}
\end{equation*}
On the other hand, by the permutation-invariance,
\begin{equation*}
E(E_n\hat X)^2=\frac1{n^2}\sum_{i,j\in[n]}E\hat X_i\hat X_j \\
=\frac n{n^2}E\hat X_1^2+\frac{n^2-n}{n^2}E\hat X_1\hat X_2.
\end{equation*}
Therefore and in view of \eqref{80} and because $|\hat X_1|\le M$, we see that
\begin{equation}
E\hat X_1\hat X_2\to0. \tag{90}\label{90}
\end{equation}
So, recalling that the $Y_i$'s are independent of the $X_i$'s, we get
\begin{equation*}
E(E_n\hat X\hat Y)^2
=\frac1{n^2}\sum_{i,j\in[n]}E\hat X_i\hat X_j\,E\hat Y_i\hat Y_j \\
=\frac n{n^2}E\hat X_1^2\,E\hat Y_1^2
+\frac{n^2-n}{n^2}E\hat X_1\hat X_2\,E\hat Y_1\hat Y_2.
\end{equation*}
Therefore and in view of \eqref{90} and because $|\hat X_1|\le M$ and $|\hat Y_1|\le M$, we see that
\begin{equation}
E(E_n\hat X \hat Y)^2\to0.
\end{equation}
Now \eqref{20b} follows by Markov's inequality. $\quad\Box$
