This problem seems to be a variant of this one, but the same argument doesn't hold here since we can't say that $f$ is determined by its values in the unit square.
I tried to reduce the problem to the original one by defining $g(z)=f(z)+iz$, which doesn't work since it gives $g(z+i)=g(z)$, $g(z+1)=g(z)+2i$. Any clues on how to approach this?
Here's an answer following a clue suggested by @GerryMyerson.
Let $f$ be an entire function as described in the question. Its derivative $g$ is also an entire function. For $g$ we have that $g(z+i)=g(z)$, $g(z+1)=g(z)$. Using the proof here we have that g is constant, which means $f$ is a linear function. Hence, $f(z)=az+b$ for some $a,b \in \mathbb C$.
Using the original constraints for $f$, we get: $az+ai+b=a(z+i)+b = f(z+i) = f(z)+1 = az+b+1 => ai=1 =>a=-i$ $az+a+b=a(z+1)+b = f(z+1) = f(z)+i = az+b+i => a=i$
Which gives $i=-i$, a contradiction.
