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Mac Lane showed that any combination of multiplications lead to the same function, if they have the same domain and codomain, via coherence. Since both $F(g \circ h)$ and $F(g) \circ F(h)$ are combinations of multiplications, they are equal via that theorem.

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So you have arrows $f:I\to[A,A']$ and $g:I\to[B,B']$. The functor we are interested in and which you called $S$ is the composite $$\mathcal V_0\times\mathcal V_0\to (\mathcal{V\otimes V})_0\to \mathcal V_0 \\ \mathcal V_0(I,[A,A'])×\mathcal V_0(I.[B,B'])\to \mathcal V_0(I,[A,A']⊗[B,B'])\to \mathcal V_0(I,[A⊗B,A'⊗B']) \\ (f,g)\mapsto (f⊗g)l^{-1}\mapsto ... 1 Actually, it might be that Mac Lane made a small error (gasp!) This error can be seen simply by seeing the domain and codomain don't match up.$$\mu^{k_1 + k_2 + ... + k_n}:c \otimes c \otimes \dots \otimes c \space (k_1+\dots+k_n \text{times}) \to c\mu(\mu^{k_1} \otimes \mu^{k_2} \otimes \dots \otimes \mu^{k_n}):(c \otimes \dots \otimes c \space (k_1 ...

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I'll use $\cal A$ to denote the category $Alg(1)$. Note that $\cal A$ is the category whose objects are endomorphisms, and it can be identified with the functor category $\mathbf{Set}^\Bbb N$, where $\Bbb N$ is the category with a single object and with natural numbers as arrows and their addition as composition. That's because $\Bbb N$ is the free category ...

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Consider the following diagram The right hand triangle commutes if every other rectangle/triangle in the diagram does. But making extensive use of the pentagonal and the triangle identities and the naturality of $\alpha$, we can show that they do. Now since $-\otimes e$ is naturally isomorphic to the identity functor, commutativity of the rightmost triangle ...

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So we take $\eta:e\to\coprod_n a^n$ to be the inclusion of $e=a^0$ into the coproduct. We want to show that $\mu(\eta\otimes1)=\lambda:e\otimes\coprod_n a^n\to\coprod_n a^n$. Consider the following commutative diagram $$\begin{matrix} & & e⊗∐_n a^n & → & ∐_m a^m⊗∐_n a^n & → & ∐_{m,n}a^m⊗a^n & → & ∐_k a^k \\ & & ... 1 If I understand correctly, you want to show that the outer pentagon of the following diagram commutes. The middle square commutes because the arrows going upwards are, on the individual summand a^k\otimes(a^l\otimes a^m), the tensor product of the inclusions i_k:a^k\to\coprod a^k,\ i_l,\ i_m, so this is just naturality of \alpha. Zooming into the ... 0 Your first map$$\coprod_{k,m,n} a^k\otimes(a^m\otimes a^n)\to Fb\otimes(\coprod_{m,n} a^m\otimes a^n)\to Fb\otimes (Fb\otimes Fb)\to^{\mu(1\otimes \mu)} Fb is given on components $a^k\otimes (a^m\otimes a^n)$ by the composition of the associators $a^k\otimes(a^m\otimes a^n)\to a^k\otimes a^{m+n}\to a^{k+m+n}$. The other map is given by the composition of ...

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