I believe there are no exterior algebras in sight. To see this, let us think through the $1$-categorical case carefully. We have a commutative ring $R$ and the ordinary category of $R$-modules $\mathrm{Mod}_R^\heartsuit$. We are considering symmetric monoidal functors $\tau_{\le k}(\Omega^\infty(S))\to \mathrm{Mod}_R^\heartsuit.$

**For $k=0$**, we are looking at symmetric monoidal functors from $\tau_{\le 0}(\Omega^\infty(S))=\pi_0(S) =\mathbf Z$ into $R$-modules, hence we are discussing cmmutative $\mathbf Z$-graded $R$-algebras. The free one, generated by $R$, is certainly the algebra $R[t^{\pm 1}]$. This *is not* the polynomial $R$-algebra $R[t]$, as claimed in the OP, which would instead be obtained if, rather than $\mathbf Z$-gradings, we were considering $\mathbf Z_{\ge 0}$-gradings.

**For $k\ge 1$**, since $\mathrm{Mod}_R^\heartsuit$ is a $1$-category, functors from $\tau_{\le k}(\Omega^\infty(S))$ are equivalent to those from $\Omega^\infty(S)$ itself. The latter is the free grouplike $\mathbb E_\infty$-space on a single generator, hence a functor $\Omega^\infty(S)$ is *equivalent* to specifying an invertible $R$-module. For any such choice of $M\in \mathrm{Pic}(R)$, the corresponding symmetric monoidal functor $F^{\Omega^\infty(S)}_M:\Omega^\infty(S)\to\mathrm{Mod}_R^\heartsuit$ will send object-wise $\mathbf Z\ni n\mapsto M^{\otimes n}$. Any symmetric monoidal functor $F:\Omega^\infty(S)\to\mathrm{Mod}^\heartsuit_R$ will be of this for, i.e. $F=F^{\Omega^\infty(S)}_M$ for some $M\in \mathrm{Pic}(R)$!

Conversely, if we were instead looking at symmetric monoidal functors $F:\mathcal F\mathrm{in}^\simeq\to\mathrm{Mod}_R^\heartsuit,$ any $R$-module $M$ would do, with the functor $F_M^{\mathcal F\mathrm{in}^\simeq}$ now going object-wise $\mathbf Z_{\ge 0}\ni n\mapsto M^{\otimes n}$. In this case, we can even see the full functoriality here: a finite set $I$ is sent to the tensor power $M^{\otimes I}.$ To compute its underlying commutative $R$-algebra, we must take the colimit $\varinjlim_{I\in \mathcal F\mathrm{in}^\simeq}M^{\otimes I}$. Using the presentation $\mathcal F\mathrm{in}^\simeq \simeq \bigoplus_{n\ ge 0}\mathrm B\Sigma_n$, we find that the underlyign commutative ring is $\bigoplus_{n \ge 0}(M^{\otimes n})/\Sigma_n = \bigoplus_{n\ge 0}\mathrm{Sym}^{\heartsuit, n}_R(M) = \mathrm{Sym}^{\heartsuit, *}_R(M)$, the usual symmetric algebra on $M$.

The Barrat-Priddy-Quillen equivalence $\Omega^\infty(S) \simeq (\mathcal F\mathrm{in}^\simeq)^\mathrm{gp}$ now implies a relationship between the functors $F^{\Omega^\infty(S)}_M$ and $F^{\mathcal F\mathrm{in}^\simeq}_M$, or more precisely, their underlying commutative $R$-algebras. Indeed, the commutative $R$-algebra $\varinjlim(F^{\Omega^\infty(S)}_M)$ is a localization of the commutative $R$-algebra $\varinjlim(F^{\mathcal F\mathrm{in}^\simeq}_M) = \mathrm{Sym}^{\heartsuit, *}_R(M)$. It is the localizations along the elements in degree one, i.e. along $\mathrm{Sym}^{\heartsuit, 1}_R(M) = M$ - we must invert all these elements to obtain the underlying commutative $R$-algebra of $F^{\Omega^\infty(S)}_M$. That is not too hard to do, but I am not aware of it having a name in general.

In the special case when $M = R$ though, things become easier! We have $\mathrm{Sym}^{*, \heartsuit}_R(R) = R[t]$, the polynomial $R$-algebra, and to invert things in degree $1$, we need just invert $t$. Hence we find that the underlying commutative $R$-algebra of $F^{\Omega^\infty(S)}_R$ is again just $R[t^{\pm 1}]$., same in the $t=\infty$ase as in the $t=0$ one.

And this is also the underlying commutative $R$-algebra of the symmetric monoidal functors $F^{\tau_{\le k}(\Omega^\infty(S))}_R$, for all $k\ge 1$. Indeed, as we discussed before, all these functors are equivalent to one another, and for $k=\infty$, we clearly get the constant functor with vale $R$. We are therefore getting the same constant functor $R$ for all $1\le k \le \infty$. The underlying commutative $R$-algebra is obtained as colimit in $\mathrm{Mod}_R^{\heartsuit}$, but since that is a $1$-category, it is oblivious to all the higher structure, and the colimit will be the same for all $1\le k\le \infty$ as well.

**Long story short:** for both $k=0$ and $k\ge 1$, we are getting the underlying commutative $R$-algebra be $R[t^{\pm 1}]$, never anything resembling exterior algebas. If you ask me, the "$1$-categorical miracle" we are running into is probably that $R[t]$ is both $\bigoplus_{n\ge 0}R$, i.e. the colimit of the constant $R$-valued functor $\mathbf Z_{\ge 0}\to\mathrm{Mod}_R^\heartsuit$, and $\mathrm{Sym}^{\heartsuit, *}_R(R) = \bigoplus_{n\ge 0} (R^{\otimes n})/\Sigma_n$, i.e. colimit of the constant $R$-valued functor $\mathcal F\mathrm{in}^\simeq\to\mathrm{Mod}_R^\heartsuit.$

That would be false if we were working in $R$-module spectra, where the $\infty$-categorical structure allows us to be more sensitive to the $\mathrm B\Sigma_n$-components of $\mathcal F\mathrm{in}^\simeq$, leading to quotients $(R^{\otimes n})_{h\Sigma_n}\simeq R_{h\Sigma_n}$ not needing to be just $R$. Indeed, the homotopy groups $\pi_i(R_{h\Sigma_n}) = \mathrm H_i(\Sigma_n; R)$ coincide with group homology, which might be non-trivial if $n$ divides the characteristic of $R$.

But even then, we have two choices:

- We follow the OP, and consider symmetric monoidal functors $\tau_{\le k}(\Omega^\infty(S))\to\mathrm{Mod}_R$. That will be some kind of thing interpolating between the localizations $R[t^{\pm 1}]$ for $k=0$, and $R\{t^{\pm 1}]$ for $k=\infty$. I don't have much insight about what to say about the intermediate stages, other than to remark that these are intermediate "localized symmetric" algebras, still not "exterior" in any way.
- But if instead we wish to compare $R[t]$ and $R\{t\}$, then we will be looking at symmetric monoidal functors $\mathbf Z\to\mathrm{Mod}_R$ and $\mathcal F\mathrm{in}^\simeq\to \mathrm{Mod}_R$ instead. Then the "$k$-truncation" idea will not be as fruitful, ad $\mathcal F\mathrm{in}^\simeq$ is a $1$-category, so we only have two options: $k=0$ for which we get $R[t]$, and $1\le k\le \infty$ for which we get $R\{t\}$.