The symmetric monoidal functor
\[ ((-)^{+},(-)^{+,\times },(-)^{+,\times }_{\mathbb {1}}) \colon (\mathsf{Sets},\times ,\mathrm{pt}) \to (\mathsf{Sets}_{*},\wedge ,S^{0}), \]
of Chapter 6: Pointed Sets, Item 4 of Proposition 6.4.1.1.2 lifts to an equivalence of categories
\begin{align*} \mathsf{CoMon}(\mathsf{Sets}_{*},\wedge ,S^{0}) & \mathrel {\smash {\overset {\scriptscriptstyle \text{eq.}}\cong }}\mathsf{CoMon}(\mathsf{Sets},\times ,\mathrm{pt})\\ & \cong \mathsf{Sets}. \end{align*}
We follow Lemma 2.4 of [PS, Coalgebras in Symmetric Monoidal Categories of Spectra].
Faithfulness
Given morphisms $f,g\colon X\to Y$, if $f^{+}=g^{+}$, then we have
\begin{align*} f(x) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f^{+}(x)\\ & = g^{+}(x)\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}g(x) \end{align*}
for each $x\in X^{+}$, and thus $f=g$, showing $(-)^{+}$ to be faithful.
Fullness
Let $f\colon X^{+}\to Y^{+}$ be a morphism of comonoids in $\mathsf{Sets}_{*}$. By counitality, the diagram commutes. If $f(x)=\star _{Y}$ for $x\neq \star _{X}$, the commutativity of this diagram then gives
\begin{align*} 1 & = \epsilon ^{+}_{X}(x)\\ & = \epsilon ^{+}_{Y}(f(x))\\ & = \epsilon ^{+}_{Y}(\star _{Y})\\ & = 0, \end{align*}
which is a contradiction. Thus $f$ is an active morphism of pointed sets, so there exists a map $f^{-}$ such that $(f^{-})^{+}=f$ (Chapter 6: Pointed Sets, Item 1 of Proposition 6.4.2.1.2).
Essential Surjectivity
Let $(X,\Delta _{X},\epsilon _{X})$ be a comonoid in $\mathsf{Sets}_{*}$. We claim that
\[ \Delta _{X}(x)=x\wedge x \]
for each $x\in X$ with $x\neq \star _{X}$. Indeed:
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Suppose that $x\neq \star _{X}$ and write $\Delta _{X}(x)=x_{1}\wedge x_{2}$.
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Since $\operatorname {\mathrm{id}}_{X}\wedge \epsilon _{X}$ is pointed, we have
\[ [\operatorname {\mathrm{id}}_{X}\wedge \epsilon _{X}](x_{1}\wedge x_{2})=\star _{X\wedge S^{0}}. \]
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The counitality condition for $\Delta _{X}$, corresponding to the commutativity of the diagram
gives
\begin{align*} x\wedge 1 & = \rho ^{\mathsf{Sets}_{*},-1}_{X}(x)\\ & = [\operatorname {\mathrm{id}}_{X}\wedge \epsilon _{X}\circ \Delta _{X}](x)\\ & = [\operatorname {\mathrm{id}}_{X}\wedge \epsilon _{X}](\Delta _{X}(x))\\ & = [\operatorname {\mathrm{id}}_{X}\wedge \epsilon _{X}](x_{1}\wedge x_{2})\\ & = \star _{X\wedge S^{0}},\end{align*}
which is a contradiction. Thus $x_{1}\neq \star _{X}$.
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Similarly, if $x\neq \star _{X}$, then $x_{2}\neq \star _{X}$.
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Next, we claim that $\epsilon _{X}(x_{2})=1$, as otherwise we would have
\begin{align*} \star _{X\wedge S^{0}} & = x_{1}\wedge 0\\ & = [\operatorname {\mathrm{id}}_{X}\wedge \epsilon _{X}](x_{1}\wedge x_{2})\\ & = [\operatorname {\mathrm{id}}_{X}\wedge \epsilon _{X}](\Delta _{X}(x))\\ & = [\operatorname {\mathrm{id}}_{X}\wedge \epsilon _{X}\circ \Delta _{X}](x)\\ & = \rho ^{\mathsf{Sets}_{*},-1}_{X}(x)\\ & = x\wedge 1, \end{align*}
a contradiction. Thus $\epsilon _{X}(x_{2})=1$.
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Similarly, if $x\neq \star _{X}$, then $\epsilon _{X}(x_{1})=1$.
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Now, since $\Delta _{X}$ is counital, we have
\begin{align*} x\wedge 1 & = \rho ^{\mathsf{Sets}_{*},-1}_{X}(x)\\ & = [\operatorname {\mathrm{id}}_{X}\wedge \epsilon _{X}\circ \Delta _{X}](x)\\ & = [\operatorname {\mathrm{id}}_{X}\wedge \epsilon _{X}](\Delta _{X}(x))\\ & = [\operatorname {\mathrm{id}}_{X}\wedge \epsilon _{X}](x_{1}\wedge x_{2})\\ & = x_{1}\wedge 1, \end{align*}
so $x=x_{1}$.
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Similarly, $x=x_{2}$, and we are done.
Next, notice that $X\cong \epsilon ^{-1}_{X}(0)\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}\epsilon ^{-1}_{X}(1)$, and let $x\in \epsilon ^{-1}_{X}(0)$. We then have
\begin{align*} [(\operatorname {\mathrm{id}}_{X}\wedge \epsilon _{X})\circ \Delta _{X}](x) & = [\operatorname {\mathrm{id}}_{X}\wedge \epsilon _{X}](x\wedge x)\\ & = x\wedge 0\\ & = \star _{X\wedge S^{0}}. \end{align*}
The counitality condition for $\Delta _{X}$ then gives $x=\star _{X}$, so $\epsilon ^{-1}_{X}(0)=\left\{ \star _{X}\right\} $. Thus we have $(\epsilon ^{-1}_{X}(1))^{+}\cong X$, and this isomorphism is compatible with the comonoid structures when equipping $\epsilon ^{-1}_{X}(1)$ with its unique comonoid structure. This shows that $(-)^{+}$ is essentially surjective.
Equivalence
Since $(-)^{+}$ is fully faithful and essentially surjective, it is an equivalence by Chapter 11: Categories, Item 1b of Item 1 of Proposition 11.6.7.1.2.
