$(\infty,1)$ 2d TFTs

As you point out, if you just look at the operad of bordisms with exactly one output disc, you get the framed $E_2$-operad (framed here means you can rotate the discs) and so the value of the circle is a framed $E_2$ algebra. (Aside: if we were working in categories an $E_2$-algebra is a braided monoidal category, while a framed $E_2$-algebra would also be balanced). Similarly it's a framed $E_2$-coalgebra. So it's natural to try to defined a framed Frobenius $E_2$-algebra to be a framed $E_2$-algebra and an $E_2$-coalgebra with some kind of "Frobenius" compatibility structure. The problem is that I can't really imagine what description of this Frobenius structure one could possibly hope for that won't make this statement just a tautology... That is, the definition of $E_2$-algebra is already highly geometric, and so one should expect the definition of Frobenius $E_2$-algebra to also be highly geometric, and hence the best way to define Frobenius $E_2$-algebra is to say its the value of the circle under a 2d TFT. One way one could hope to make this approach less tautological would be to say what kind of more "combinatorial" description of $E_2$-algebras you'd like the putative combinatorial description of Frobenius $E_2$-algebras to extend. For example, you could define an $E_2$ algebra to be an $A_\infty$-algebra in $A_\infty$ algebras to give a Stasheff-like combinatorial description. But you'd still need to also put in the balanced structure somehow.

Perhaps better would be to take an approach that generalizes a different definition of Frobenius algebra. Instead of saying "algebra and coalgebra with Frobenius condition" one can instead say "commutative algebra A with trace such that Tr(xy) is non-degenerate." This has a natural generalization:

Def: A framed Frobenius $E_2$-algebra is, a framed $E_2$-algebra $A$ together with an $\mathrm{SO}(2)$-invariant map $\mathrm{Tr}: A\rightarrow \mathbf{1}$ such that $X \boxtimes Y \mapsto \mathrm{Tr}(X \cdot Y)$ is the evaluation of a duality between $A$ and itself.

So now we have the following precise question:

Question: Is the map from the space of 2-dimensional TFTs valued in a symmetric monoidal $(\infty,1)$-category equivalent to the space of framed Frobenius $E_2$-algebras under the functor which evaluates the circle?

I don't know how to prove this and am not totally sure if it's true, but since duals are defined like adjoints, you can hope that the "uniqueness of coherent adjoints" lets you recover the TFT just from the framed Frobenius $E_2$-algebra structure.

I'm not even sure if this is true in the case where the target is the $(2,1)$-category of categories... My understand there from Bakalov-Kirillov is that this is asking about the classification of modular functors, which are classified by weakly rigid balanced tensor categories. But this seems to me to be saying that the trace has to be given by $\mathrm{Hom}(-,\mathbf{1})$, and I don't see how that matches up with my definition here of a Frobenius $E_2$-algebra. Maybe I'm just misunderstanding Bakalov-Kirillov's assumptions though.

Edited answer: As I said in the comment, I think that the structure you're looking for is that of a cyclic algebra over the framed little disc operad, which should coincide with the notion suggested by Noah, ie a framed $E_2$-algebra equipped with a non-degenerate invariant trace compatible with the structure.

Recall that a PROP is a symmetric monoidal ($\infty$-)category which is equivalent to one with has $\mathbb{N}$ as a set of object, with tensor product given by addition. An algebra over a PROP is just a symmetric monoidal functor out of it. A PROP is cyclic if the object $1$ (not to be confused with the unit !), hence every other object, is self-dual, and the evaluation is a symmetric pairing (ie formally the canonical pivotal structure you get is the identity). If $P$ is a PROP then you can look at the endomorphism operad of $1$. This operation has a left adjoint giving the free PROP on an operad: roughly just declare that $Hom(n,1)$ is the $n$th operation space of your operad, and extend the obvious way.

If the PROP is cyclic, ie $1$ is equipped with a symmetric non-degenerate pairing, then the endomorphism operad of 1 has a canonical cyclic (in fact automatically modular) structure the usual way, using the isomorphisms $$Hom(n,1) \cong Hom(n \otimes 1,0)\cong Hom(n+1,0)$$ given by this pairing to extend the $S_n$-action to an $S_{n+1}$-action. Hence you get a cyclic operad, and this operation again has a left adjoint: take the same construction as before, and then there is a unique way of adding an evaluation and coevaluation, in such a way that the induced cyclic structure on the endomorphism operad of $1$ is the cyclic structure which was given in the first place.

Remark Note that "being cyclic" is an additional structure on an operad, and the same holds for cyclic algebras. On the other hand cyclicity for PROP is "built in", a cyclic PROP is just a special kind of PROP, which is why I find those convenient.

So: the familiar category whose objects are disjoint unions of oriented discs, and morphisms smooth oriented embedding, is nothing but the PROP associated with the framed little disc operad. This operad is equivalent to the operad of genus 0 Riemann surfaces, equivalently the disc PROP/category is equivalent to the PROP with objects disjoints unions of circles, and morphisms disjoint union of genus 0 surfaces with exactly one output boundary (given a configuration of discs in a larger disc, simply cut the smaller discs out).

Now the operad of genus 0 Riemann surfaces has an obvious cyclic structure. I claim that the cyclic PROP associated to it, is nothing but $Cob(2)$: basically you just formally add the macaroni to make $S^1$ self-dual and then all possible compositions. The as in the classical case all relations formally follow from the genus 0 ones, and from properties of the trace/pairing. Note that the fact that I can switch between pairing and trace follows from the fact taht everybody in sight is unital, ie in all those PROP's I have a morphism $0 \rightarrow 1$ given either by embedding of the empty disc and by the "cap" respectively.

Warning I haven't actually checked this carefully. I'm fairly sure it was explained, in the slightly different language of modular operads, in an old paper of Getzler but I haven't been able to find it.