Let $k(x,y) \in C^0([0, \infty) \times [0, \infty), \mathbb R)$. Assume that the following linear operator \begin{align*} K : L^2([0, \infty)) &\rightarrow L^2([0, \infty)) \\ f &\mapsto K(f) := \int_0^{\infty} k(x,y)f(y)dy \end{align*} is well-defined and bounded.
Can we deduce that $k(x,y) \in L^2([0, \infty) \times [0, \infty))$ ?
No, you can't. For example let $g(x)=e^{-\frac 1 2|x|}$ and $k(x,y)=g(x-y)$. Then $K(f)=g\ast f\in L^2$ with $\|K(f)\|_2\leq \|g\|_1\|f\|_2$ by Young's inequality (set $f(x)=0$ for $x<0$ for the convolution). But $$ \int_0^\infty\int_0^\infty k(x,y)^2\,dx\,dy=\int_0^\infty \left(\int_0^y e^{x-y}\,dx+\int_y^\infty e^{y-x}\,dx\right)\,dy=\int_0^\infty(2-e^{-y})\,dy=\infty. $$