Imagine a red-black tree. Is there always a sequence of insertions and deletions that creates it?
The insert and delete operations in a red-black tree include the balancing needed to maintain the red-black properties.  

The problem with non(left- or right-) leaning red black trees is that there are multiple ways to restore the red-blackness after the basic delete or insert.
It is not the insert or the delete that transforms the tree, but the rebalancing and rotation that happens afterwards to preserve/restore the red blackness of the tree.  

The basic description of the red-black tree does not prescribe which of the possible routes to take.
It may not be possible to figure out how to exactly reconstruct a given red black tree, because the rebalancing need not be deterministic.    

This has been 'solved' with left-leaning red black trees.
There is only one way the balancing is done. So any given leaning red black tree can be reconstructed using inserts and deletes, because the rebalancing/rotations are done in a specific deterministic way.    

This does not mean that left-leaning RB-trees are better or more efficient, what they gain on one hand by using deterministic balancing rules, they lose on the other by more complex balancing code.

As per @Anton's comment:
There is an algorithm which uses the operation insert and delete to construct a valid red-black tree consisting only of black nodes. It uses $(h+2)⋅2^{h}−1$ insertions/deletions to create a tree of height $h$. First, we can create a perfectly balanced red-black tree in breadth-first manner using $2^{h+1}−1$ insertions, then using $h∗2^{h-1}$ insertions and the same amount of deletions repaint it into a completely black tree. The trick here is to move up $h$ times the lowest red layer up the tree until it reaches the root. 

I think a complete balancing algorithm like Day-Stout-Warren would be more efficient though.