As a discerning dissenting voice, let me say that it ismight 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.
It is allThere are two definitions in your definition, doc!!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. Off courseAn 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 Fred and Dylan rightly point outjust a category. Together with the first definition of abelian category, this enrichmentmakes the statement true. An alternative is inherent in yourto think of a category structurethat cannot be abelian (essentially the negative of the second definition). This choice would make the statement false.