Firstly, I'm going to assume you want $x'$ to be nonzero. Secondly, do you want $x$ and $xx'$ to also both be non-zero?

If so, then there is a silly counterexample -- take $n=1$, and take $x$ to be $id - t$ where $t$ is the permutation $1\to 2\to \dots \to m\to 1$. Then $x$ lies in the augmentation ideal of $F[S_{m+n}]$ and so if $xx'$ lies in $\theta^*(F[S_1]) = F$ then it would have to be zero.

I strongly suspect that one could play similar tricks with higher values of $n$.

NOTE: This answer was given in response to an earlier version of the question above, and so probably seems completely irrelevant if you can't see the old version. (More strikethrough and less deletion in editing, please?)

Firstly, I'm going to assume you want $x'$ to be nonzero. Secondly, do you want $x$ and $xx'$ to also both be non-zero?

If so, then there is a silly counterexample -- take $n=1$, and take $x$ to be $id - t$ where $t$ is the permutation $1\to 2\to \dots \to m\to 1$. Then $x$ lies in the augmentation ideal of $F[S_{m+n}]$ and so if $xx'$ lies in $\theta^*(F[S_1]) = F$ then it would have to be zero.

I strongly suspect that one could play similar tricks with higher values of $n$.