Are regular languages closed against an intersection that keeps words with the same number of ones? How can we show that the class of regular languages is closed under the following operation? 

Let $L_1$ and $L_2$ be laguages over $\Sigma=\{0, 1\}$. 

The operation is: $$\{x \in L_1 \mid \text{ for some } y \in L_2, \text{ strings } x \text{ and } y \text{ contains equal numbers of } 1s \}$$ 

Is the only way to show this to create a NFA of the new language? 

I have done the following: 

Let $M_{L_1}=(Q_{L_1},\Sigma ,\delta_{L_1},q_{L_1},F_{L_1})$ the DFA that recognizes $L_1$ and $M_{L_2}=(Q_{L_2},\Sigma ,\delta_{L_2},q_{L_2},F_{L_2})$ the DFA that recognizes $L_2$. 

Let $M=(Q,\Sigma ,\delta,q,F)$ the NFA that recognizes the operation. 

$Q=Q_{L_1}\times Q_{L_2}, \ \ q=(q_{L_1},q_{L_2})$. 

Is this correct? Which is the transition function?
Let 
$$
  L = \{ x \in L_1 \mid \text{there exists }y \in L_2 \text{ such that } |x|_1 = |y|_1\}
$$
Let $\pi: \Sigma^* \to a^*$ be the homomorphism defined by $\pi(u) = a^{|u|_1}$.
I claim that
$$
  L = L_1 \cap \pi^{-1}(\pi(L_2))
$$
Since regular languages are closed under homomorphisms, inverses of homomorphisms and intersection, it will follow that $L$ is regular.

Proof of the claim.
\begin{align}
  L_1 \cap \pi^{-1}(\pi(L_2)) &amp;= \{ x \in L_1 \mid \pi(x) \in  \pi(L_2) \} \\
  &amp;= \{ x \in L_1 \mid \text{there exists }y \in L_2 \text{ such that } \pi(x) = \pi(y)\}\\
  &amp;= \{ x \in L_1 \mid \text{there exists }y \in L_2 \text{ such that } |x|_1 = |y|_1\} \\
  &amp;= L
\end{align}