I prove that what you're trying to show is not true first in a simple but non general way and then in a more abstract and general way.
Simple proof:
Take $x_1,x_2 \in [a,b]$ and $y_1,y_2 \in [c,d]$, then consider the operator $T \; : \; C^k([a,b] \times [c,d]) \to \mathbb{R}^{2 \times 2}$ that maps $f$ in the matrix
$$T(f) = \begin{pmatrix}
f(x_1, y_1) & f(x_1, y_2) \\
f(x_2, y_1) & f(x_2, y_2)
\end{pmatrix}$$
consider the polynomial
$$g(x,y) = \frac{(x - x_2)(y-y_2)}{(x_1-x_2)(y_1-y_2)} + \frac{(x - x_1)(y-y_1)}{(x_2-x_1)(y_2-y_1)}$$
then
$$T(g) = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}$$
and so $\det(T(g)) = 1 \neq 0$. On the other hand the function $f \mapsto \det(T(f))$ is continuous function with respect to the topology of $C^k$ but for any function of the form $f(x,y) = f_1(x)\cdot f_2(y)$ we have
$$\det(T(f)) = \det\bigg(\begin{pmatrix} f_1(x_1) \\ f_1(x_2) \end{pmatrix} \cdot \begin{pmatrix} f_2(y_1) & f_2(y_2) \end{pmatrix} \bigg) = 0
$$
therefore if $S := \{ f_1(x)\cdot f_2(y) \; : \; f_1 \in C^k([a,b]), f_2 \in C^k([c,d])$ we have that $\det(T(f)) = 0$ for all $f \in S$, and so by continuity $\det(T(f)) = 0 \forall f \in \overline{S}$ but this means $g \not\in \overline{S}$ since $\det(T(g)) \neq 0$
For $L^1(\mathbb{R}^2)$ a similar argument works defining
$$T(f) = \begin{pmatrix}
\iint_{X_1 \times Y_1}f & \iint_{X_1 \times Y_2}f \\
\iint_{X_2 \times Y_1}f & \iint_{X_2 \times Y_2}f
\end{pmatrix}$$
for some disjoint sets of finite positive measure $X_1,X_2,Y_1,Y_2 \subset \mathbb{R}$.
Abstract proof:
Let $X$,$Y$ be banach spaces and consider the Banach Space tensor product $X \otimes Y$, consider the space $S := \{ x \otimes y \; : \; (x,y) \in X \times Y\}$ then $\overline{S} \neq X \otimes Y$ whenever $\min(\dim(X),\dim(Y)) \geq 2$. To prove this take some distinct $\ell_1,\ell_2 \in X'$ and $x_1,x_2 \in X$ such that
$$ \begin{pmatrix}
\ell_1(x_1) & \ell_2(x_1) \\
\ell_1(x_2) & \ell_2(x_2)
\end{pmatrix} = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix} $$
Define in a similar way $h_1,h_2 \in Y'$ and $y_1,y_2 \in Y$. Now I can define on $X \times Y$ the map $\ell_i \otimes h_j (x \otimes y) := \ell_i(x) \cdot h_j(y)$, I can then extend this by linearity and continuity over all $X \otimes Y$, this defines the function $\ell_i \otimes h_j$, so I can define the operator
$$T(z) = \begin{pmatrix}
(\ell_1 \otimes h_1)(z) & (\ell_1 \otimes h_2)(z) \\
(\ell_2 \otimes h_1)(z) & (\ell_2 \otimes h_2)(z)
\end{pmatrix}$$
then $\det(T(x_1 \otimes y_1 + x_2 \otimes y_2)) = 1$ but for $x \otimes y$ we have
$$\det(T(x \otimes y)) = \det\bigg(\begin{pmatrix} \ell_1(x) \\ \ell_2(x) \end{pmatrix} \cdot \begin{pmatrix} h_1(y) & h_2(y) \end{pmatrix} \bigg) = 0
$$
and so by continuity $\det(T(z)) = 0 \forall z \in \overline{S}$ but then $x_1 \otimes y_1 + x_2 \otimes y_2 \not\in \overline{S}$
