Let's define measures $\mu^+$ and $\mu^-$ on Lebesgue-measurable subsets of $[0,1]\times [0,1]$ by
$\mu^+(E)=\int_E f^+(x,y)dxdy$
and
$\mu^-(E)=\int_E f^-(x,y)dxdy,$
where $f^+$ and $f^-$ are the positive and negative parts of $f$.
It's a simple exercise (check!), using boundedness of $f$ (a condition we could relax), that these are bona fide measures. Consider the family of boxes
$\mathcal{A}=\{[a,b]\times[c,d]:0\leq a\leq b \leq 1,~ 0\leq c\leq d\leq 1\}.$
This is a $\pi$-system, since the overlap of two boxes is again a box. $\mu^+$ and $\mu^-$ agree on this $\pi$-system, and hence (see these notes or try yourself using Dynkin's Lemma) on the $\sigma$-algebra generated by the $\pi$-system. So they agree on $\mathcal{B}([0,1]\times[0,1])$. Both measures are $0$ on sets of measure $0$, so in fact they agree on the completion of $\mathcal{B}([0,1]\times[0,1])$, $\text{LEB}([0,1]\times[0,1])$.
So if $E\in \text{LEB}([0,1]\times[0,1])$,
$\mu^+(E)-\mu^-(E)=\int_E f(x,y)dxdy=0.$
Since $f$ is Lebesgue-measurable, we may consider the set on which $f$ is positive. The integral over this set is $0$, so, for any $n\in\mathbb{N}$, $f$ is less than $1/n$ a.e. (why?), so $f$ is non-positive a.e.. Then we may similarly show that it's non-negative, and thus $0$ a.e..