Determine reaction concentration at chemical equilibrium with multiple reactions
Answer already found by myself. Short conclusion:

To calculate variations of the input concentrations at chemical equilibrium 2 major steps are needed.

1) Calculate the equilibrium constants $K_x$
for every reaction $x$ at given concentrations. Solve that system of x equations (details in 2nd edit) with a numerical solver of your choice. 

EDIT: No numerical solver is needed, since the equations simplify to:

$\ce{K_1 - \frac{CN}{C*N} = 0}$

$\ce{K_2 - \frac{CO}{C*O} = 0}$

which is very interesting, since this simplifies the whole problem a lot. It also means that at chemical equilibrium a reaction does not care if a reactant is depleted via another reaction e.g., is bound within another product, or if the (initial) concentration of that reactant is just lowered to achieve the same amount in equilibrium. No reaction describes the connection between $\ce{CO}$ and $\ce{CN}$ e.g., $\ce{CO + N &lt;=&gt; CN + O}$, so basically the concentration of $\ce{CO}$ and $\ce{CN}$ ARE DECOUPLED (looking at the decoupled equilibrium constants), although they seem to be coupled via their common reactant ($\ce{C}$) e.g., changing $\ce{N}$ (or $\ce{O}$), changes the amount of produced $\ce{CN}$ (or $\ce{CO}$), changes the amount of $\ce{C}$ and further the amount of the other product $\ce{CO}$ (or $\ce{CN}$) .

This gives rise to a new question: How would a 'forced coupling' between $\ce{CO}$ and $\ce{CN}$ via $\ce{CO + N &lt;=&gt; CN + O}$ influence the concentration of both products? 

Since we are at chemical equilibrium and have already taken 'indirect coupling via $\ce{C}$' into account, I guess that introducing a triangular-like reaction scheme would only introduce a redundant equation and won't give new information. Maybe I will test for that and share my results.

EDIT: Testing showed no difference in using a triangular-like reaction scheme (including the reaction above). As stated above, the 'new' reaction/equation does not contain new information. The 'new' equation can be written as a linear combination of the other 2 reaction equations. Therefore, one of these 3 reaction equations can be canceled/eliminated since 1) we need $n$ equations for $n$ variables and 2) an equation consisting of a linear combination of other equations cancels out (leads to a zero result) when solving the system of equations.

This means we only need to consider (in this case) the 3 equations for mass/element conservation and 2 reaction equations for the 5 variables (concentration of 3 reactants and 2 products).



2) Adjust/variate the input abundance and use another system of $z$
equations (conversation of mass/element and reaction equations - details in the edit of the question) to calculate the new concentrations at equilibrium.

Note: Make sure to always have $n$ equations for $n$ variables and don't mess them up.