Leftist heap - determining time complexity
For the time complexity of merging, JeffE already pointed out in the comments that $O(\log(n_1n_2)) \in O(\log(n_1+n_2))$.

For arbitrary deletions note the following:


When we merge the two subtrees of the deleted node, the root of the resulting tree will have an s-value that is at least the s-value of the deleted node -1.
Thus, the s-value of any updated node will not be decreased by more than 1. (There is no constant bound for how much it can increase.)
Thus, if we update the s-value of a left child in and that value was strictly larger than the s-value of the right child, the s-values of the parent and all nodes above the parent will not change.
If the s-value of the left child and right child are equal, exchanging them still gives a valid leftist heap.
Thus, the number of updated nodes is bound by the length of a right flank in a leftist tree, which in turn is bound by $O(\log n)$.