Matching Algorithm - How to maximize matched quantity with unique matching rules?
The problem you are describing is a special case of the (weighted) 3-dimensional matching problem.1

Unlike regular 2-dimensional matching, maximum cardinality 3-dimensional matching is NP-hard, which means that weighted 3-dimensional matching is NP-hard. Therefore, solving this problem exactly in polynomial time doesn't seem feasible, unless the specific case you have here is easier than the general problem.



This special case seems easier than the general problem. My idea is to represent it as a minor modification to the regular 2d maximum weight matching problem. Note that for each of your combinations of 3 items, there is at least one item type that is not present in another combination (e.g. $E$ is only present in the combination $(A,D,E)$), I will call such an item type conflict-free. First, select a conflict free type for each combination of $3$ types: Here, I pick $E$ for $(A,D,R)$, $F$ for $(B,C,F)$, and $H$ treat items of such type differently than the others.

More concretely, for any combination of items $x,y,z$ of type $X,Y,Z$, where $Z$ is the chosen conflict-free type, we add only the edge $(x,y)$ of weight $f(q(x), q(y),q(z))$, where $q(i)$ denotes the quantity of item $i$ and $f$ is a function that determines the value of a matching based on the weight (including the type is of course also possible). Combinations with only $2$ types can be modeled as usual. Solving the maximum weight matching problem on this graph (which in your case is bipartite) provides the maximum sum of values that can be made.

For example, on the input $[A:5,\ A:4,\ D:6,\ B:3,\ E:5]$, the graph will look like the following:



Red edges and their weights indicate the maximum matching. Note that the item of type E is not part of the graph, but only influences the weight of the two edges between types A and D. So the approach works for this case.

However, if we have items of type E with different quantity, we the weights of the edges will depend on which edges are selected. To see this, take the input $[A:5,\ A:4,\ D:6,\ D:4,\ B:3,\ E:5,\ E:1]$. We now get the following graph:



The edges between type $A$ and $D$ now no longer have a simple value as their weight, but a list of values. This is because each edge has to use one of the two $E$'s, but the same one cannot be reused. This means in particular that the blue matching has a value of $12+3=15$ instead of $12+12=14$, because the item $E$ of quantity $5$ can only be used once. Therefore, it is the red matching of value $21$ that is optimal.

I do not know whether the standard algorithms for weighted bipartite matching can deal with this modification on the weights or whether there we can make a modification to the standard algorithm to handle this. 



1: Possible matches of the form $X+Y\leftrightarrow Z$ for items $x, y, z$ can be represented by the 3d-edge $(x,y,z)$ of weight the minimum of the quantities of the items $x,y,z$. Matches of the form $X\leftrightarrow Y$ for items $x,y$ can be represented by the 3d-edge $(x,y,d)$ of weight minimum of the quantities of items $x,y$ and a dummy vertex $d$ that is in no other 3d-edge.