David Roberts mentioned in his comments the relationship
K-theory : spin group
TMF : string groups
Let me recommend the first 6 pages of my unfinished article
for a uniform construction of $SO(n)$, $Spin(n)$, and $String(n)$,
which suggests the existence of a similar uniform construction of
$H\mathbb R$, $KO$ (or $KU$), and $TMF$.
You'll see that von Neumann algebras appear in the construction.
More precisely, von Neumann algebras appear in the definition of conformal nets. The latter are functors from 1-manifolds to von Neumann algberas.
For a summary of the conjectural relationship between conformal nets and $TMF$, have a look at page 8 of
this other paper of mine (joint with Chris Douglas).
Loop groups yield non-trivial examples of conformal nets.
Those conformal nets are related (conjecturally) to equivariant $TMF$.
The details of the comparison are treated in detail in the original ABGHR paper (and then unfortunately split in half across two papers in the updated version), so I'll just try to give a sketch of what's going on.
Thom spaces
Given any type of bundle $E \to B$ we can view this as a diagram of spaces indexed by $B$. The homotopy colimit of this diagram in the homotopy theory of spaces is just $E$ itself (which makes sense- we are 'summing up over $B$').
If $E \to B$ has a section $\infty: B \to E$, then this supplies each fiber with a basepoint. The homotopy colimit over $B$ of this diagram of pointed spaces can't be $E$, because it has a whole $B$-family of basepoints. So we need to identify those all together and take the cofiber $B \to E \to E/B$. The homotopy colimit is then $E/B$.
Example. Given a pointed spherical fibration, $E \to B$, with fibers $S^n$ (pointed!), this procedure gives the Thom space. So we learn that: the Thom space of a pointed spherical fibration is the homotopy colimit taken in the homotopy theory of pointed spaces of the family of spheres indexed over the base.
Example. Given an unpointed spherical fibration $E \to B$, with fibers $S^{n-1}$, we can form the fiberwise suspension by taking the cofiber $S^{n-1} \to CS^{n-1}$, fiberwise. This produces a pointed spherical fibration, and we can then take the Thom space as above. It's not hard to see that collapsing the section at $\infty$ of this suspended fibration gives the same answer as taking the mapping cone of the projection for the original fibration, whence the connection with your first definition.
Example. If $B = BG$ is the classifying space for a group, then the 'homotopy colimit indexed by $BG$' is also called the 'homotopy orbits for the action of $G$', since we may identify a bundle $E \to BG$ with a homotopy coherent action of $G$ on the fiber over a fixed basepoint of $BG$. So, in this case, the Thom space is of the form $S^n_{hG}$ where $G$ acts in some way on $S^n$. This includes the universal example, when $G = \mathrm{hAut}(S^n)$, the space of homotopy automorphisms of $S^n$ (i.e. the degree $\pm 1$ components of $\Omega^nS^n$ when $n>0$).
Thom spectra
For bundles of spectra it is nice to take the 'diagrams indexed by $B$' point of view as a definition. So a bundle is given by a map $B \to \{\text{space of spectra equivalent to }S^0\}=\mathrm{BGL}_1(S^0)$. This, in particular, defines a diagram of spectra and its homotopy colimit is the Thom spectrum. Since $\Sigma^{\infty}_+$ commutes with taking homotopy colimits (it's a left adjoint), we see that taking the suspension spectrum of a Thom space gives the Thom spectrum of the bundle obtained by taking fiberwise suspension spectra.
Now, in the case that $B = BG$, then we may identify functors $BG \to \mathsf{Sp}$ with modules over the $\mathbb{E}_1$-ring $S^0[G] := \Sigma^{\infty}_+G$. (This is a spectral version of the relationship between $G$-modules and $\mathbb{Z}[G]$-modules for an ordinary group). So we have:
spherical fibrations over $BG$ $\iff$ a coherent action of $G$ on the sphere spectrum $S^0$ $\iff$ an $S^0[G]$-module structure on $S^0$.
Under this correspondence, taking homotopy colimits corresponds to the construction on modules $M \mapsto M \otimes_{S^0[G]}S^0$ where we use the augmentation. (This is analogous to the statement that the coinvariants of a $G$-module are computed via a similar tensor product in the classical setting).
Thus, in the case $B = BG$, we may compute Thom spectra as $S^0 \otimes_{S^0[G]} S^0$ where the left $S^0$ has an interesting module structure, and the second $S^0$ is acted on through the augmentation.
Of course, the action of $G$ factors through the largest action of all of $\mathrm{GL}_1(S^0)$, so we can write this as
$S^0 \otimes_{S^0[G]}S^0 = S^0 \otimes_{S^0[\mathrm{GL}_1(S^0)]} S^0[\mathrm{GL}_1(S^0)] \otimes_{S^0[G]} S^0$.
In the special case when $G \to \mathrm{GL}_1(S^0)$ is 'normal', ie arises as the fiber of an $\mathbb{E}_1$-map $\mathrm{GL}_1(S^0) \to H$ (which happens int he infinite loop space context if we take $H = \Omega^{\infty}Cj$ in your notation), then we may simplify the right hand side of the tensor product as
$S^0 \otimes_{S^0[\mathrm{GL}_1(S^0)]} S^0[\mathrm{GL}_1(S^0)] \otimes_{S^0[G]} S^0=S^0 \otimes_{S^0[\mathrm{GL}_1(S^0)]} S^0[H]$.
So that's the relationship between the two definitions you write down.
Best Answer
The answer to the question
is No in general. As you say, $A(X)$ is determined by the homotopy type of $\Sigma^\infty \Omega X_+$ as an associative (or $A_\infty$) ring spectrum, and this ring spectrum does not uniquely determine $X$, even if $X$ is simply-connected.
For example, suppose $Z$ is a pointed space. Let $T(Z)$ be the free associative $S$-algebra generated by $Z$. I.e., $$T(Z)=\Sigma^\infty S^0\vee \Sigma^\infty Z \vee (\Sigma^\infty Z)^{\wedge 2} \vee \cdots .$$
If $Z$ is connected, there is an equivalence of associative ring spectra $$ T(Z) \simeq \Sigma^\infty \Omega\Sigma Z_+.$$ The equivalence is a version of the classical James splitting. It is induced by a map of spectra $\Sigma^\infty Z \to \Sigma^\infty \Omega\Sigma Z_+$, extended to a map of ring spectra $T(Z)\to \Sigma^\infty \Omega\Sigma Z_+$ using freeness.
It follows that if $X$ and $Y$ are connected spaces such that $\Sigma X$ and $\Sigma Y$ are not equivalent, but $\Sigma^\infty X$ and $\Sigma^\infty Y$ are equivalent, then there is an equivalence $A(\Sigma X)\simeq A(\Sigma Y)$ providing a counterexample.
A couple of comments:
It is well-known that there exist non-isomorphic groups $G$ and $H$ such that the group rings $\mathbb Z[G]$ and $\mathbb Z[H]$ are isomorphic. There are even examples with finite $G$ and $H$. One may wonder if for some of these examples the spherical group rings $\Sigma^\infty G_+$ and $\Sigma^\infty H_+$ are equivalent as associative ring spectra. If yes, then $BG$ and $BH$ would provide another counterexample.
In general one can have non-equivalent ring spectra that have equivalent $K$-theories. For example, I believe that if $P$ and $Q$ are Morita equivalent in a suitable sense, then $K(P)\simeq K(Q)$. Can there be two spaces $X$ and $Y$ such that $\Sigma^\infty \Omega X_+$ and $\Sigma^\infty \Omega Y_+$ are not equivalent as ring spectra, but have equivalent categories of modules (in a strong enough sense to induce equivalence of $K$-theories)? It seems far fetched, but I don't know how to exclude this possibility. Added later: A paper by Roggenkamp and Zimmerman gives an example of two groups $G$ and $H$ for which the rings $\mathbb Z[G]$ and $\mathbb Z[H]$ are not isomorphic, but Morita equivalent. It follows that the Quillen $K$-theory of these rings is isomorphic. One may ask whether the $K$-theory spectra of $\Sigma^\infty G_+$ and $\Sigma^\infty H_+$ are equivalent as well.