commutative "weakly" Frobenius algebras and 2d TQFT Fix a field $k$.  A classic result written up carefully by Abrams in the article "Two-Dimensional Topological Quantum Field Theories and Frobenius Algebras"
  says that there is a bijective correspondence between finite dimensional commutative Frobenius algebras and 2d TQFTs in the sense of Atiyah.
Suppose we do not require that $A$ is finite dimensional but still require that the pairing is nondegenerate i.e. the map $A \to Hom(A,k)$ is injective. I think the center  $C_k[G]$ of any group ring $k[G]$ satisfies this but I haven't checked it. Is there a similar description in terms of 2D surfaces?
 A: Let's first think about 1-dimensional TQFTs.  As is well-known, these correspond exactly to finite dimensional vector spaces as follows. The positive point is assigned to some vector space V, the negative point is assigned to some vector space W, and the "cap" gives a pairing between them, the "cup" gives an element $\sum v_i \otimes w_i$.  The zig-zag relations tell you that the pairing is non-degenerate, that the cup is the copairing $\sum e_i \otimes e^i$, and that the vector space is finite dimensional.  (This is a great exercise to work out if you haven't done so before.)  The value of the circle is the dimension of the vector space.
Ok, now we want to find a way to "break" this to generalize to infinite dimensional vector spaces.  The only good way I can see to do this is to replace the bordism category with the "non-compact" or "punctured" bordism category whose morphisms are 1-dimensional bordisms such that every connected component has non-trivial incoming boundary.  This keeps the pairing, but gets rid of the co-pairing.  A functor from the punctured 1-dimensional bordism category to vector spaces is a 1-dimensional punctured TQFT.  Now an infinite dimensional vector space gives you a perfectly good punctured TQFT, however these aren't the only ones.  In fact any pair of vector space V and W with a pairing between them should do, and there's no reason that pairing needs to be non-degenerate in any sense whatsoever.
You could also ask about unoriented 1-dimensional TQFTs and those would correspond to a vector space V with a (possibly degenerate) symmetric billinear form.
Ok, now let's turn to 2-dimensions.  Note that the above discussion is highly relevant because any 2-dimensional TQFT gives you a 1-dimensional one by "dimensional reduction."  That is if Z is a 2-dimensional TQFT then we get a 1-dimensional one via $Z(M \times S^1)$.  In particular, as pointed out in comments the value of the torus in your 2-dimensional TQFT is $\dim Z(S^1)$ and so you have to break either the pairing or copairing in some way.
Again we can look at punctured bordisms and punctured TQFTs.  This means we have perfectly good bordisms for pants, copants, downwards macaroni, and cap.  However, we no longer get upwards macaroni or cup.  Translating across to algebra, we still get a multiplication, comultiplication, and trace, and many compatibility rules like associativity.  But we no longer have a unit or a copairing.  In particular, infinite dimensional Frobenius algebras give such examples, but they're not the only examples because there's no requirement of being unital or of the pairing being non-degenerate.
Punctured TQFTs (more commonly called "non-compact TQFTs") are studied by Costello and others in the context of "Calabi-Yau algebras."
