A complex equation I want to solve the following equation:
$g(s)f(s)=0$
where $f$ and $g$ are defined in the complex plane with real values and they are not analytic. 
My question is: 

If I assume that $f(s)≠0$, can I deduce that $g(s)=0$ without any further complications?

I am a little confused about this case: if $f=x+iy$ and $g=u+iv$, then $fg=ux-vy+i(uy+vx)$ and $fg$ can be zero if $ux-vy=0,uy+vx=0$ without the implications: $x=y=0,u=v=0$

First, note that your question is really just about individual complex numbers, not about complex-valued functions.
Now, as you note, if $(x + i y)(u + i v) = 0$ then this implies that $x u - v y = x v + u y = 0$.  However, the only way these equations can hold is if either $x + i y = 0$ or $u + i v = 0$.  
Multiplying the first equation through by $v$ and the second by $u$, we have $x u v - v^2 y = 0$ and $x u v + u^2 y = 0$.  Subtracting, $(u^2 + v^2)y = 0$.  Since $u, v, y \in \mathbb{R}$, this means that either $y = 0$ or $u = v = 0$.  In the latter case, we have $u + i v = 0$.  In the former, since $y = 0$ we have $x (u + i v) = 0$, so $x u = x v = 0$.  Either $x = 0$, in which case $x + i y = 0$, or $u = v = 0$, in which case $u + i v = 0$ again.
