Is this a known question about the expression of a function on $\Bbb R^2$ as an infinite sum of products?

In "Representation of functions of two variables as sums of rectangular functions, I", Roy O. Davies shows that under the continuum hypothesis every function has a representation of this form.

More precisely, he shows that we can get a representation with the additional property that for all $x,y$, the sum has only finitely many non-zero terms. Conversely, he proves that if there is a representation with this additional property for $(x,y) \mapsto e^{xy}$, then the continuum hypothesis holds.

In his survey Set Theoretic Real Analysis(p. 18), Krzysztof Ciesielski mentions this problem (and a lot of interesting similar results), it seems that it is open whether or not the continuum hypothesis is equivalent to the existence of weak representations.

At least, assuming sufficient large cardinals, it's consistent that there is a function without any such representation.

Note that any function of the form $\sum_{n=1}^\infty g_n(x) h_n(y)$ is measurable with respect to the product $\sigma$-algebra $\mathcal{P}(\mathbb{R}) \otimes \mathcal{P}(\mathbb{R})$. However, as explained in product of power sets, assuming large cardinals, it's consistent that $\mathcal{P}(\mathbb{R}) \otimes \mathcal{P}(\mathbb{R}) \ne \mathcal{P}(\mathbb{R}^2)$. If that's so, then if $A \in \mathcal{P}(\mathbb{R}^2) \setminus (\mathcal{P}(\mathbb{R}) \otimes \mathcal{P}(\mathbb{R}))$, the function $f(x,y) = 1_A(x,y)$ has no representation as a sum of products.

This may be huge overkill, and there could still be an easy ZFC counterexample, but at least it suggests that people can stop looking for a ZFC proof that every function has a sum-of-products representation.

Under the continuum hypothesis (or various other axioms, e.g. MA), on the other hand, we actually do have $\mathcal{P}(\mathbb{R}) \otimes \mathcal{P}(\mathbb{R}) = \mathcal{P}(\mathbb{R}^2)$. Together with the multiplicative system theorem (or functional monotone class theorem or similar), that would imply something close to the opposite answer: there is no nontrivial vector subspace of functions on $\mathbb{R}^2$ that contains all the products $g(x) h(y)$ and is closed under pointwise convergence. (Roughly speaking, you should allow not only pointwise limits of sums of products, but limits of limits of limits of ..., iterated up to $\omega_1$-many times.)

This is not meant as an answer, but as a too-long-for-comment response to a comment.

I claim that $f(x, y) = \begin{cases} 1 & x = y \\ 0 & \text{else} \end{cases}$ can be represented in this form.

Consider the Cantor set $\mathcal{C}$ as the space of all (one-way infinite) bitstrings. Let $a: \mathbb{R} \rightarrow \mathcal{C}$ be any bijection. Let $q_i$ be an enumeration of finite bit-strings (e.g., $q_0$ is the blank string, $q_1 = \text{"0"}$, $q_2 = \text{"1"}$, $q_3 = \text{"00"}$, …). Then let $g_1 = h_1 \equiv 1$; let $g_{2i}(x) = \begin{cases} 1 & \text{$a(x)$ starts with $q_i+\text{"0"}$} \\ 0 & \text{else,} \end{cases}$ $h_{2i}(y) = \begin{cases} -1 & \text{$a(y)$ starts with $q_i+\text{"1"}$} \\ 0 & \text{else,} \end{cases}$ $g_{2i+1} = h_{2i}$, and $h_{2i+1} = g_{2i}$ (where the "+" denotes concatenation of strings, e.g. $"101" + "0" = "1010"$).

If $x = y$, then only $g_1$, $h_1$ affect the sum, so the sum is $1$. If $x \ne y$, then there is a unique longest bitstring that both $a(x)$, $a(y)$ start with; then either $g_{2i} h_{2i}$ or $g_{2i+1} h_{2i+1}$ will "cancel" the $+1$ of $g_1 h_1$. The other $g_j h_j$ will be $0$ at $(x, y)$, so the total will be $0$. Therefore, the sum is $1$ on the diagonal, and $0$ elsewhere.