Let $(\Omega,\mathcal{F},P)$ be a probability space. Suppose $f_n,g_n, n\in \mathbb{N}$ are sequences of functions on this space such that their product $f_ng_n$ converges weakly in $L^2$ to $h$, say. Moreover suppose $g_n$ converges strongly to 1 (you may take $g_n$ to be $\chi_{E_n}$ where $E_n \uparrow \Omega$). Then, does $f_n$ converge weakly, and in particular, to $h$?
Edited later: If $f_n$ lie in a particular subspace of $L^2$ (say if each $f_n$ is measurable wrt to a sub-$\sigma$-algebra $\mathcal{G}$), then does $h$ have to lie in that subspace?
In general, this is false (see below), but if $(f_n)_n$ is bounded,it is true, since we have for every $f \in L^2$ that
$$\langle f_n , f\rangle = \langle f_n g_n , f\rangle + \langle (f_n (1-g_n), f\rangle ,$$ where the first summand on the right converges to $\langle h,f\rangle$ and where $$ |\langle f_n (1-g_n), f\rangle \leq \|f_n\|_2 \|(1-g_n)f\|_2 \to 0, $$ since $(f_n)_n$ is bounded, since $g_n = \chi_{E_n}$ and by dominated convergence.
Finally, in general the claim is false, as we can see by taking $h=0$ and $f_n$ an arbitrary $L^2$ function vanishing on $E_n^c$ with $\|f_n\|_2 \to\infty$. Then $f_n g_n =0=h$ for all $n$, but $(f_n)_n$ cannot converge weakly, since it is not bounded.
EDIT: Here is the solution to the more general case.
First, for arbitrary (closed) subspaces, what you want does not hold. For example, since we are on a probability space, $L^2 \hookrightarrow L^1$, so that $V := \{f \in L^2 \,\mid\, \int f \, d\mu =0 \}$ is a closed subspace of $L^2$. But we can arrange $f_n \equiv 1$ on $E_n$ and choose $f_n \equiv c_n$ on $E_n^c$, so that $f_n \in V$. Nevertheles, we will have $f_n g_n \to 1$ in $L^2$, with $1 \notin V$.
But for the case of functions which are measurable w.r.t. a certain sigma-subalgebra, the answer turns out to be affirmative. In fact, this holds for any subspace $V \leq L^2$ which is closed in $L^2$ with respect to almost everywhere convergence.
By Mazur's Lemma (cf. https://en.wikipedia.org/wiki/Mazur%27s_lemma), there is a function $N : \Bbb{N} \to \Bbb{N}$ and a bunch of real numbers $\alpha_n^{(k)}$ for $n \in \Bbb{N}$ and $k \in \{n, \dots, N(n)\}$ satisfying $\sum_{k=n}^{N(n)} \alpha_n^{(k)} = 1$ and such that $h_n := \sum_{k=n}^{N(n)} \alpha_n^{(k)} f_n g_n$ converges strongly to $h$ as $n \to \infty$.
Now, by switching to a subsequence, we get almost everywhere convergence $h_{n_\ell} \to h$ almost everywhere, i.e. everywhere on $\Omega \setminus \Theta$ for a null-set $\Theta \subset \Omega$.
We will show below that this even implies $F_n \to h$ almost everywhere for $F_n := \sum_{k=n}^{N(n)} \alpha_n^{(k)} f_n$. Now, clearly $F_n \in V$ (this only requires $V$ to be convex, i.e. we don't need it to be a subspace) and thus $h \in V$, since we assume $V$ to be closed in $L^2$ with respect to almost everywhere convergence.
To see $F_n \to h$ on $\Omega \setminus \Theta$, note that any $x \in \Omega \setminus \Theta$ satisfies $x \in E_n$ for all $n \geq n_0$ for some $n_0 =n_0(x)$ by your assumption. Hence, $g_n(x) = 1$ for $n \geq n_0$. Since the sum defining $h_n$ begins at $n$, we get $$ h(x) \leftarrow h_n (x) = \sum_{k=n}^{N(n)} \alpha_n^{(k)} f_n (x) g_n (x) = \sum_{k=n}^{N(n)} \alpha_n^{(k)} f_n (x) =F_n (x) $$ for $n \geq n_0$.