Kruskal is O(E log E); your derivation is right. You could also say O(E log V) because E <= V * V, so log(E) <= 2 log(V) (I don't know why I remember that, other than that I think a prof put that on an exam at one point...)
Since |V| > |E|+1, we prefer a tight upper bound with V terms instead of E terms.
|E| <= |V|²
. log |E| < log |V|²
. log |E| < 2 log |V|
. running time of MST-KRUSKAL is: O(E log V)