Is every element of a complex semisimple Lie algebra a commutator?

Yes, this holds for all complex simple Lie algebras. A reference is Theorem $A$ in the article On commutators in a simple Lie algebra. The result can be extended to simple Lie algebras over more general fields.

1) In addition to Brown Theorem above, it is worth noting the following "1.5" generators of such Lie algebras.

Theorem: Let L be a simple Lie algebra over an infinite field k of char 0 (or char not 2,or 3). Then L=[L, a] +[L, b] for some a, b in L. In particular, every element of L is a sum of at most 2 commutators.

Reference: G. Bergman, N. Nahlus, Homomorphisms of infinite direct product algebras especially Lie algebras, J. Algebra, 2011, pp. 67-104. (Thm. 26)

In fact, there are some explicit "1.5" generators over C (complex field) which can be found in some papers by Panyushev.

2) Back to "every element of L must be a single commutator", It is indeed true in COMPACT real semisimple Lie algebras

References: 1) For a proof using Kostant Convexity Theorem, see Appendix 3 in the book by Karl Hofmann & Sidney Morris, The Lie Theory of connected Pro-Lie groups, 2007. In fact, this proof was communicated to the authors by Karl-Hermann Neeb.

2) For another "implicit" proof, see Thm. 3.4 in "Some Questions about Semisimple Lie Groups Originating in Matrix Theory", Canad. J. Math. Vol. 46 (3), 2003, pp. 332-343 by Dragomir Z. Dokovi´c and Tin-Yau Tam. Their proof uses Dynkin diagrams

3) Another reference is by D. Akhiezer, On the Commutator Map for Real Semisimple Lie Algebras, 2015, arXiv:1501.02934

4) The question whether "every element of L must be a single commutator for any Real semisimple Lie algebra L" is an open question.