As a discerning dissenting voice, let me say that it might be true, depending on your definitions.

Take abelian $A$. Let $B:=A$ with forgotten enrichment in abelian groups. Then the identity functor is equivalence, but $B$ is not abelian because it is not even additive.

There are two definitions in play. The first is *abelian category*. The standard (ncatlab or Wikipedia) definition of abelian category asks for the category to be [pre-additive][1], which requires enrichment in abelian groups that can be forgotten. An alternative definition, mentioned in other answers, is from Freyd's book: there abelian is an inherent property rather than an additional structure. This definition would make the statement false.

The second definition is  *non-abelian category*. There is no accepted definition. One possibility is to think of this as just a category. Together with the first definition of abelian category, this makes the statement true. An alternative is to think of a category that cannot be abelian (essentially the negative of the second definition). This choice would make the statement false.


  [1]: https://ncatlab.org/nlab/show/pre-additive+category