The skew right unitor of the right tensor product of pointed sets is the natural transformation
whose component
\[ \rho ^{\mathsf{Sets}_{*},\rhd }_{X} \colon X\rhd S^{0}\to X \]
at $\webleft (X,x_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$ is given by the composition
\begin{align*} X\rhd S^{0} & \cong |X|\odot S^{0}\\ & \cong \bigvee _{x\in X}S^{0}\\ & \rightarrow X, \end{align*}
where $\bigvee _{x\in X}S^{0}\to X$ is the map given by
\begin{align*} \webleft [\webleft (x,0\webright )\webright ] & \mapsto x_{0},\\ \webleft [\webleft (x,1\webright )\webright ] & \mapsto x \end{align*}
for each $x\in X$.
Firstly, note that, given $\webleft (X,x_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$, the map
\[ \rho ^{\mathsf{Sets}_{*},\rhd }_{X} \colon X\rhd S^{0}\to X \]
is indeed a morphism of pointed sets as we have
\[ \rho ^{\mathsf{Sets}_{*},\rhd }_{X}\webleft (x_{0}\rhd 0\webright )=x_{0}. \]
Next, we claim that $\rho ^{\mathsf{Sets}_{*},\rhd }$ is a natural transformation. We need to show that, given a morphism of pointed sets
\[ f\colon \webleft (X,x_{0}\webright )\to \webleft (Y,y_{0}\webright ), \]
the diagram
commutes. Indeed, this diagram acts on elements as
and
and hence indeed commutes, showing $\rho ^{\mathsf{Sets}_{*},\rhd }$ to be a natural transformation. This finishes the proof.