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 $(X,x_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$ 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 $(X,x_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$, 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}(x_{0}) & = 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
\[ f\colon (X,x_{0})\to (Y,y_{0}), \]
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.