Im asking in general but to give a little context, i'm studying group schemes and have been trying to prove that the composition morphism $m:G\times G \rightarrow G$ is uniquely determined by the composition law $f\cdot g = m\circ (f\times g)$ for all $f,g:T\rightarrow G$.
My only guess is that indeed $m$ is unique because of its value on all pairs $(f,g)$.
So your category $\mathcal{C}$ has finite products and is locally small since you have a magma structure on your Hom-spaces.
For this proof, I need the functoriality of the composition law, namely for $h:T' \to T$, $(f \circ h)\cdot (g\circ h) = (f\cdot g) \circ h$.
Consider the representable presheaf $G(-) = \hom(-,G) = h_G$. The Yoneda embedding is define to be the functor $$\begin{array}{rrcl} h: &\mathcal{C} &\to &\hat{\mathcal{C}} \\ &X&\mapsto&h_X \end{array}$$ where $\hat{\mathcal{C}}$ is the category of presheaves over $\mathcal{C}$. You can find in the literature that $h$ is fully faithfull, but for it to be essentially surjective, you need to have a small category (hopefully we don't need it here).
We are looking to find $m \in \hom(G \times G,G)$ and all we know is the presheaf morphism between the two representables (so a natural transformation) $\phi: \hom(-,G \times G) \to \hom(-,G)$ defined by $$\forall T \in \mathcal{C}, \quad \begin{array}{rrcl} \phi_T: &\hom(T,G \times G) &\to &\hom(T,G) \\ &(f,g)&\mapsto&f\cdot g \end{array}$$ This is where we need the functoriality of the composition law (otherwise this would not be a natural transformation).
But $\phi \in \hom_\hat{\mathcal{C}}(h_{G \times G},h_G)$ which is isomorphic to $\hom_\mathcal{C} (G \times G, G)$ via $n \mapsto n_*$. Hence there exists a unique $m: G \times G \to G$ such that $m_* = \phi$, which gives you the desired morphism.
Now for any other morphism $m'$, you can check that $m_ = m_*'$ hence by faithfulness, $m=m'$.