The statement is false. The standard definition of "reductive" for a finite dimensional Lie algebra $\mathfrak{g}$ over an arbitrary field of characteristic 0 is given in a number of equivalent ways by Bourbaki in Chapter 1 (1960) of their treatise on Lie groups and Lie algebras: section 6, no. 4-5. By definition, $\mathfrak{g}$ is *reductive* provided its adjoint representation is semisimple (= completely reducible). Typical equivalent conditions: the derived algebra is semisimple; or $\mathfrak{g}$ is the direct sum of a semisimple and an abelian Lie algebra; or the solvable radical equals the center.
As a consequence, a finite dimensional representation of a reductive Lie algebra is semisimple iff the center acts by semisimple endomorphisms. (An abelian Lie algebra need not be represented in that way.)

Some of this is set up as an exercise at the end of Section 6 in my Springer graduate text (1972); see also Proposition 19.1.