I believe so (caveat: this is not something I've every thought about before).
I'm going to rewrite the $p_x$ from your question as $p_{xy}$. So, we have
$$
p_{xy}=|\langle x|U_1|0^n\rangle|^2\ |\langle y|U_2|0^n\rangle|^2
$$
Note that this is two independent probabilities $p_x$ and $p_y$.
Now, the probability that $p_{xy}>\alpha'$, which we write as $\mathbb{P}(p_{xy}>\alpha')$ certainly satisfies
$$
\mathbb{P}(p_{xy}>\alpha')>\mathbb{P}(p_{x}>\sqrt{\alpha'})^2
$$
because if both the independent probabilities $p_x$ and $p_y$ are bigger than $\sqrt{\alpha'}$, the probability of the produce being larger than $\alpha$ (which includes some events not counted in the independent case) is certainly larger.
Now, from the paper you cite (theorem 5, $\epsilon=0$), we know
$$
\mathcal{P}\left((p_x>\frac{\alpha}{N}\right))\geq\frac{(1-\alpha)^2}{2}
$$
So, equate
$$
\sqrt{\alpha'}=\frac{\alpha}{N}
$$
and follow through to find the constant $\beta'$ such that
$$
\mathbb{P}(p_{xy}>\alpha')>\beta',
$$
the definition of being anti-concentrated.