As I wrote in the title, my question is:
Is the resultant face of joining an even number of odd faces an even face?
For instance, we can see that the statement is true for the following graph: we are joining four triangular faces and the new one is a hexagon.
I believe that the statement is true, and I tried to prove it by using the fact that when you add a new face, you are losing an edge of the face that you are including. However, this fact is not true in general since you can join new faces and lose more than one edge.
Consider the planar graph formed by joining those $2k$ odd faces. It has $2k+1$ faces total: the last face is the face we formed.
The sum of all $2k+1$ face lengths is even, because that sum is always twice the number of edges. However, the sum of the first $2k$ face lengths is also even, because we have added together $2k$ odd numbers. Therefore the $(2k+1)^{\text{th}}$ face must be an even face.
(You could also try to adapt your inductive argument. The idea you need is that when you join a new odd face that shares $j$ edges with the previous faces, you actually lose $2j$ edges: $j$ old ones, and $j$ edges you would otherwise have gotten from the new face.)