Which $z(x,y)$ are possible solutions from the following equations?
\begin{align} S(z) &= xP(y)+Q(y),\\ T(z) &= y M(x)+N(x),\\ G(z) &= y+x F(z) \end{align}
Where the functions $S,P,Q,T,M,N,G$ and $F$ can all be chosen arbitrarily.
I have found $z=x+y$ so far, by setting $P(y)=1,Q(y)=y,M(x)=1,N(x)=x,F(z)=1,S(z)=z,T(z)=z,G(z)=z,$
and $z=y/x$ by setting $P(y)=1/y,Q(y)=0,M(x)=1/x,N(x)=0,G(z)=0,S(z)=1/z,T(z)=z,F(z)=-z$
Is there a general way to find all solutions $z(x,y)$?
First of all, note that not all arbitrary choices for the given functions $ S $, $ T $, $ \dots $ end up in a situation where there exists a solution $ z ( x , y ) $ for the system consisting of the equations $$ S ( z ) = x P ( y ) + Q ( y ) \text , \tag 0 \label 0 $$ $$ T ( z ) = y M ( x ) + N ( x ) \text , \tag 1 \label 1 $$ $$ G ( z ) = y + x F ( z ) \text . \tag 2 \label 2 $$ For a very trivial example, take $ T ( z ) = F ( z ) = G ( z ) = 0 $, $ N ( x ) = 1 $ and $ S $, $ P $, $ Q $ and $ M $ arbitrary. Then \eqref{2} shows that the system can only be solved if we restrict the domain of the variable $ y $ to contain nothing but $ 0 $, but even then, \eqref{1} shows that no solution exists. More nontrivial examples can be given.
Second of all, as already indirectly mentioned, the question as is posed at the moment is vague because no domain for the variables $ x $, $ y $ and $ z $ is explicitly considered. In the last paragraph, I've assumed that these variables can take real values, but possibly not all of them. This was motivated by your second example where $ x $, $ y $ and $ z $ cannot be equal to zero, as they appear in denominator of some fractions. So, you need to specify the following things before the problem completely makes sense: You need to specify domains and codomains for all the given functions $ S $, $ T $, $ \dots $, and also for the function $ z ( x , y ) $ that we're looking for. These domains and codomains need to satisfy some properties for the question to make sense. For example we need to have $ \operatorname {codom} z \subseteq \operatorname {dom} S \cap \operatorname {dom} T \cap \operatorname {dom} G \cap \operatorname {dom} F $, because otherwise $ S ( z ) $, $ T ( z ) $, $ G ( z ) $ and $ F ( z ) $ don't make sense. You also need to specify which values of $ x $ and $ y $ need to satisfy the equations \eqref{0}, \eqref{1} and \eqref{2}. These should also be in line with the previous specifications, so that for example for all permitted $ x $ and $ y $ and each $ w \in \operatorname {codom} F $, $ y + x w $ be a member of $ \operatorname {codom} G $.
Third of all, assuming we've specified everything needed to make the question well-posed, and assuming the given functions $ S $, $ P $, $ \dots $ are chosen in a way that the system has a solution, there are cases where you have infinitely many solutions, which are not easily classifiable and nice like the ones in your own examples. For example, take $ P $, $ T $, $ M $, $ N $ and $ F $ to be constantly zero, $ S ( z ) = G ( z ) = z ^ 3 - z $, and $ Q ( y ) = y $. Take all the domains and codomains to be $ \mathbb R $ (with the exception of $ \operatorname {dom} z $ which should be $ \mathbb R ^ 2 $), and demand that \eqref{0}, \eqref{1} and \eqref{2} be satisfied for all real numbers $ x $ and $ y $. \eqref{1} will be satisfied trivially, and \eqref{0} and \eqref{2} will both become $$ z ( x , y ) ^ 3 - z ( x , y ) = y \text . \tag 3 \label 3 $$ The function $ S $ is surjective, so for every given $ y $ you can choose a value for $ z $ satisfying \eqref{3}. But $ S $ is not injective (as for example we have $ S ( - 1 ) = S ( 0 ) = S ( 1 ) = 0 $), and thus for some values of $ y $ (exactly those with $ - \frac 2 { 3 \sqrt 3 } \le y \le \frac 2 { 3 \sqrt 3 } $) we have more than one choice. We can combine this with the fact that for a fixed $ y $ we're free to choose the value of $ z ( x , y ) $ differently for different $ x $. This way, we end up with an infinite number of solutions for what the function $ z $ can be, and infinitely many of them won't be of a very nice form.
The above explanations show that different specifications of the problem may have very different properties, from having no solutions at all, to having a unique nice solution (like your own examples), to having a wildly infinite set of ugly solutions. So, one should not hope to find a uniform way of dealing with the general problem, and should try to solve specific instances of the problem apart from each other, in different ways.
I would like to add that in cases where $ S $ or $ T $ is injective, \eqref{0} or \eqref{1} alone will show the only possible solution, $ z ( x , y ) = S ^ { - 1 } \big( x P ( y ) + Q ( y ) \big) $ or $ z ( x , y ) = T ^ { - 1 } \big( y M ( x ) + N ( x ) \big) $. In that case, one can take the possible solution above, put it in the other equations and check whether they are satisfied. In case they are both satisfied you've found the solution you sought, and in case one of them is not satisfied you've shown that there is no solution. This is in fact what has happened in the case of your own examples, and that's why you could find the unique solution.