Since this is tagged complex analysis:
(1) Such a function does not exist if you assume it is complex analytic on its domain and the various integrals evaluate to non-zero values.
Proof: apply Cauchy's integral theorem to the following contour: starting from the origin, go horizontally to $(1,0)$, go up to $(1,1)$, and travel back down diagonally to $(0,0)$.
(2) An analytic function with your properties must also be periodic. That is: $f(0,y) = f(1,y)$ and $f(x,0) = f(x,1)$.
Proof: consider the contour from the origin to $(1,0)$ to $(1,y)$ to $(0,y)$ and back to the origin. This shows that $\int_0^y f(0,t) dt = \int_0^t f(1,t)dt$ for every $t$. Apply the fundamental theorem of calculus.
From (2), apply Rademacher's characterisation of analytic functions (see Theorem 10 in this), you have that a periodic extension of $f$ is an analytic function on the complex plane, which immediately implies that $f\equiv 0$.
So, to summarize: if $f$ is a function with your properties, which in addition is analytic in $(0,1)\times(0,1)\subset\mathbb{C}$ and is continuous up to the boundary, then $f\equiv 0$.