The skew left unitor of the right tensor product of pointed sets is the natural transformation
whose component
\[ \lambda ^{\mathsf{Sets}_{*},\rhd }_{X} \colon X \to S^{0}\rhd X \]
at $\webleft (X,x_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$ is given by the composition
\begin{align*} X & \rightarrow X\vee X\\ & \cong |S^{0}|\odot X\\ & \cong S^{0}\rhd X, \end{align*}
where $X\to X\vee X$ is the map sending $X$ to the second factor of $X$ in $X\vee X$.
Firstly, note that, given $\webleft (X,x_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$, the map
\[ \lambda ^{\mathsf{Sets}_{*},\rhd }_{X} \colon X \to S^{0}\rhd X \]
is indeed a morphism of pointed sets, as we have
\begin{align*} \lambda ^{\mathsf{Sets}_{*},\rhd }_{X}\webleft (x_{0}\webright ) & = 1\rhd x_{0}\\ & = 0\rhd x_{0}.\end{align*}
Next, we claim that $\lambda ^{\mathsf{Sets}_{*},\rhd }$ 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 hence indeed commutes, showing $\lambda ^{\mathsf{Sets}_{*},\rhd }$ to be a natural transformation. This finishes the proof.
