Translation surfaces with constant zero curvature

A translation surface is a surface that is the sum of two planar curves contained in orthogonal planes. Thus a parametrization of the surfaces is $$X(u,v)=(x,0,f(x))+(0,y,g(y))$$ where $f$ and $g$ are smooth functions defined in some intervals of ${\mathbb R}$. We are interesting in the translation surfaces with $H=0$ or $K=0$ on the whole surface. 

For the mean curvature, it is immediate that $H=0$ is equivalent to $$\frac{f''(x)}{1+f'(x)^2}+\frac{g''(y)}{1+g'(y)^2}=0.$$ Then necessarily we have that $$\frac{f''(x)}{1+f'(x)^2}=-\frac{g''(y)}{1+g'(y)^2}=c$$ for some real number $c$. If $c=0$, then $f''=g''=0$, obtaining $f(x)=ax+b$, $g(y)=cy+d$ and $z=ax+cy+b+d$, that is, the surface is a plane. If $c\not=0$, integrating $f$ and $g$ we obtain $$f(x)=-\frac{1}{c}\log\cos(cx+m),\ g(y)=\frac{1}{c}\log\cos(cy+n),\ m,n\in{\mathbb R}.$$ Thus we write $$z=\frac{1}{c}\log\left(\frac{\cos(cy+n)}{\cos(cx+m)}\right).$$ This surface is called the Scherk's surface. In order to study the domain of the $z(x,y)$, we take $c=1$ and $m=n=0$. Then $$z=\log\left|\frac{\cos(y)}{\cos(x)}\right|.$$ Then the domain is 

$$(x,y)\in(-\frac{\pi}{2},\frac{\pi}{2})\times (-\frac{\pi}{2},\frac{\pi}{2}).$$

It is clear that in the sides of this square, the function $z=z(x,y)$ takes $\infty$ or $-\infty$ values, as it is shown in the next picture.

If we now study translation surfaces with $K=0$, then this identity is equivalent to 

$$f''g''=0.$$

Then $f''=0$ identically or $g''=0$ identically. Without loss of generality, we suppose $f''=0$, that is, $f(x)=ax+b$ for some numbers $a,b$. Then the surface writes as $z=ax+g(y)+b$ or in terms of $X$, $$X(x,y)=x(1,0,a)+(0,y,g(y)).$$ This surface is a ruled surface whose base curve is any curve as $\alpha(y)=(0,y,g(y))$ and the rulings as parallel to the direction $(1,0,a)$. In the picture we consider $a=0$ and $g(y)=\sin(y)$.