0. Clarification: I am not sure whether the green connections between the convex polygons are provided as part of the input or whether the program has to determine the appropriate green connections as shortest paths between polygons itself. I just noticed that your image is missing a green edge at the top right corner (that would be part of outer border if allowed).
1. Solution: Always choose the next edge. If your input specifies which green lines are allowed and which not, then you simply traverse the boundary of the outside component by finding one start point (e.g. by taking the polygon corner with lowest x-coordinate), ordering the green edges that depart from the polygon of this current point in counter-clock-wise order and choosing the edge immediately following the current point in CCW order. Now, take the other end of that edge as "current point" and repeat the same method to find the next edge... until you return to the first polygon.
2. Solution: Start with the convex hull. If you allow all possible green lines, you don't need them as input. The convex hull of all polygon edges is a first approximation to your solution (details of finding --- see further below): The convex hull contains real polygon edges (let's think of them as "black": They are part of your final solution, already in correct order) and edges that connect polygons (Let's think of them as red: They need to be replaced by green edges and possibly parts of other polygons).
Completing the 2nd solution: divide and conquer: Now, we need to replace each red edge by a combination of green and black edges. We just focus on one red line at a time (and apply the same method for each red line that we may have).
So we have one red line that contains two polygons that have one green line (the shortest connection between them) --- the four edges of these two lines define a quadrangle. If none of the other polygon corners are in this quadrangle, you are done: replace the red edge by the green edge and any black edges on the polygons to get to the connecting points.
But if you find polygon in the quadrangle, select from them the closest to the red edge. Move the red edge towards that point --- so that the new point cuts the red edge into two red edges. These two new red edges replace the one old red edge: apply this method recursively to both of them. Their corresponding quadrangles are much smaller and contain less polygon edges.
As you keep applying this divide and conquer method you will eventually end up with no remaining red edges (because you loose one red edge each time you find an empty quadrangle).
Convex hull: This is a difficult problem in n dimensions but easy in two dimensions: If you search the net or browse SO, you surely find a better solution than I can think of right now, but here is one that comes to mind (again: divide and conquer): Find one point A with maximal x-coordinate and one point B with minimal x-coordinate, connect them by two directed "blue" edges: A->B and B->A. Divide your points into two sets: those at the right hand side of the edge A->B and those on the left hand side (which really is the right hand side of B->A). We repeatedly replace each blue edge until we find the convex hull:
Take one blue edge A->B and look at the points its right hand side. If there are none, then this blue edge is really black (part of your solution). Otherwise, take the point C furthest to the right of the blue edge and replace the edge A->B by two blue edges A->C and C->B. Divide the points that were at the right hand side of A->B into those that are at the right hand side of A->C, those that are at the right hand side of C->B and those that are neither (they are ignored). Repeat until all blue edges are replaced.