Big caveat that I'm not a probabilist, and your answer really doesn't include much cryptography, so might be better suited for asking a probabilist somewhere (say on math.se or something).
As mentioned in the comments, this is easily false. Let $P_n, Q_n$ both be distributed as any symmetric distribution, and let $R_n\sim \{-1,1\}$ be uniform.
Define the joint distributions $P_n\times R_n$ and $Q_n\times R_n$ as follows --- the marginals on both $X$ and $Y$ are fixed as above, but
$$\Pr[(P_n\times R_n)\in E] = \Pr[(P_n\times R_n)\in E\mid R_n = b]\Pr[R_n = b].$$
Now, as we are discussing symmetry, write $X = X_1\cup X_{-1}$.
Assume the symmetry swaps these two components.
We now define the conditional distributions
$$\Pr[(P_n\times R_n)\in E\mid R_n = b] = \begin{cases}0 & E\cap X_{b}\neq \emptyset \\
2\Pr[P_n(E_X)]&\text{else}
\end{cases}.$$
This is to say the conditional distribution is defined such that a random variable with $R_n = b$ is in $X_b$, e.g. the components $P_n, R_n$ are "perfectly correlated". For $Q_n$, do the same, but reverse the roles of $X_1, X_{-1}$, e.g. have $Q_n, R_n$ be "perfectly anti-correlated".
It is straightforward to see these random variables have identical marginals, and are therefore perfectly indistinguishable (and statistically indistinguishable as well).
It is also straightforward to see that the joint distributions $P_n\times R_n$ and $Q_n\times R_n$ have disjoint supports, so
$$0 = \Delta(P_n, Q_n) \leq \Delta(P_n\times R_n, Q_n\times R_n) = 1,$$
and therefore they are not randomly statistically indistinguishable.
Note that if you assume $P_n, R_n$ are independent (in your language $E$ factors as $E_X\times E_Y$ I think), the answer is easily true.
As a sketch of the proof, by the data-processing inequality we have that $\Delta(P_n, Q_n) \geq \Delta(f(P_n), f(Q_n))$ for any randomized $f$, including $f : X\to X\times Y$ that samples $R_n$ independently, and outputs $f(x) = (x, R_n)$.
This isn't what you asked, but its still useful to note.
