Yes, every finitely presented category with a decidable word problem is an AQL schema. Here is the AQL schema for {A, B, f:A->B, g:B->A, f.g=id}:

schema S = literal : empty {
entities A B
foreign_keys f : A -> B g : B -> A
path_equations f.g = A

There's a lot more you can do, like add attributes and layer the schema over a type side so that you can do things like express equations involving addition, subtraction, etc.

By decidable word problem, I mean that there exists a decision procedure to decide if two morphisms are equivalent according the the path equations - in general, this problem is undecidable (it contains the word problem for groups). But AQL uses a variety of techniques from automated theorem proving to construct such procedures.

If you look at Conal's definition of category, a category is a Haskell type constructor of arity two, along with two polymorphic haskell operations:

class Category k where
id ::a‘k‘a

that means that Conal's definition of category encompasses only those categories who objects are exactly the Haskell types (such as the category of Haskell types and functions, namely (->)). But the category / AQL schema S has only two objects, A and B, neither of which are Haskell types. So there's no direct way to make S into an instance of Conal's category type class.

If there's a way around the issue I described above, I would very much like to learn about it.