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, 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.

with additional structuresatisfying some property, then the question does not make any sense, because $\mathcal B$ is not given any structure. Bugs Bunny seems the only non-crazy person around. $\endgroup$ – Joël Mar 29 at 0:31yes, inconsistent with your options, which I think everyone else agrees about. $\endgroup$ – Kevin Carlson Mar 29 at 1:01