Let $\epsilon^{A}_B$ be the map $\eta_{\frac{B}{A},A,B}^{-1}(1_{\frac{B}{A}}) $, or in other words, let $\epsilon^{A}$ be the counit of the adjunction $? \otimes_A\dashv \frac{??}{A}$, and similarly for $B,C$. Then the inverse of $\eta_{A,B,C}$ is the map $g\mapsto \epsilon^{B}_C \circ (g\otimes B)$. Moreover, the inner composition
$$\frac{C}{B}\otimes\frac{B}{A}\to\frac{C}{A}$$
is the image of the composite
$$\frac{C}{B}\otimes\frac{B}{A}\otimes A\stackrel{\frac{C}{B}\otimes \epsilon_{B}^{A}}{\longrightarrow} \frac{C}{B}\otimes B \stackrel{ \epsilon_{C}^B}{\longrightarrow} C$$
under the bijection $\eta_{\frac{C}{B}\otimes\frac{B}{A},A,C}$.
As a consequence, and because of the naturality of $\eta$, we find that
\begin{align}\bullet_{A,B,C}\circ (\widehat{\psi}\otimes\widehat{\varphi}) & = \eta_{\frac{C}{B}\otimes\frac{B}{A},A,C}\left(\epsilon^B_C\circ \left(\frac{C}{B}\otimes\epsilon^{A}_B\right) \right) \circ (\widehat{\psi}\otimes \widehat{\varphi})\\
& = \eta_{I\otimes I,A,C}\left(\epsilon^B_C\circ \left(\frac{C}{B}\otimes\epsilon^{A}_B\right) \circ (\widehat{\psi}\otimes \widehat{\varphi}\otimes A)\right) \\
& = \eta_{I\otimes I,A,C}\left(\epsilon^B_C\circ \left(\widehat{\psi}\otimes (\epsilon^{A}_B \circ ( \widehat{\varphi}\otimes A))\right) \right) \\
& = \eta_{I\otimes I,A,C}\left(\epsilon^B_C \circ \left(\widehat{\psi}\otimes (\varphi\circ \lambda_A) \right) \right) \\ &
= \eta_{I,A,C}\left(\epsilon^B_C \circ (\widehat{\psi}\otimes B) \circ (I\otimes (\varphi\circ \lambda_A)) \right) \\
& = \eta_{I\otimes I,A,C}\left(\psi\circ \lambda_B \circ (I\otimes (\varphi\circ \lambda_A)) \right)\\
& = \eta_{I\otimes I,A,C}\left(\psi\circ \varphi\circ \lambda_A \circ \lambda_{I\otimes A} \right) \\
& = \eta_{I\otimes I,A,C}\left((\psi\circ \varphi)\circ \lambda_A \circ (\lambda_{I}\otimes A) \right)\\
& = \eta_{I,A,C}\left((\psi\circ \varphi)\circ \lambda_A \right)\circ \lambda_{I} \\ & = \widehat{\psi\circ \varphi} \circ \lambda_I\end{align}
(with some associators missing, but it should work).
Best Answer
If $\mathcal{V}$ is closed as symetric monoidal category, this means that for any $V$ in $\mathcal{V}$, the functor $X \mapsto V\otimes X$ has a right adjoint : $Y \mapsto \{V,Y\}$, i.e. for every $V$, every $X$ and every $Y$ there is an object $\{V,Y\}$ and a natural isomorphism, $$ \eta_{X,Y}^V : Hom_{\mathcal V} (V\otimes X, Y) \to Hom_{\mathcal V}(X,\{V,Y\}). $$ You can then see $\{V,Y\}$ as an internal Hom object to $\mathcal V$ i.e. $\mathcal V$ is self enriched :
Remark : If you have a monoid $A$ in such a category consider $M$ and $N$ two left $A$-modules, then we can give $\{M,N\}$ the structure of a left $A$-module via $A\otimes \{M,N\} \to \{M,N\}$ defined as the image under $\eta$ of the morphism $A\otimes \{M,N\} \otimes M \to N$ defined as $\alpha_N \circ (id_A \otimes ev_{M,N})$ where $\alpha_N$ is the left action of $A$ on $N$, similar to the left $A$-module structure on the abelian group of morphisms of abelian groups between two left $A$-modules for a (non commutative) ring $A$, $a\cdot f : x \mapsto a\cdot f(x) $, you act ouside of $f$.
Given a monoid $A$ in $\mathcal V$, you can consider the following $\mathcal V$-enriched category : one object, and $A$ is the monoid of endomorphisms of that object. Call that category $B A$. Consider the category $Fun^\mathcal{V}(BA,\mathcal V)$ of enriched functors with $\mathcal V$ with its selfenrichement. What are these ? well it should be a map $F : ob(BA) \to ob(\mathcal V)$ between the class of objects, but since there is only one object in $BA$, this is the choice of some object $M$ in $\mathcal V$, and morphisms in $\mathcal V$, $F_{x,y}: Hom^{\mathcal V}_{BA}(x,y) \to \{F(x),F(y)\}$, but again since there is only one object in $BA$ this is just a morhphism of monoids $A \to \{M,M\}$. With the help of $\eta$ you can transform this to a classical left action of $A$ on $M$, i.e. a map $A\otimes M \to M$ that satisfies the axioms of left modules.
The category of $\mathcal V$-enriched functors between two $\mathcal V$-enriched categories $C$ and $D$ is also $\mathcal V$-enriched : Consider two functors $F,G : C\to D$, then $Nat^\mathcal{V}(F,G)$ is defined as the enriched end (so as a limit, here you need the completness asumption in $\mathcal V$): $$ \int_{c \in C} Hom^\mathcal{V}_D(F(c),G(c)). $$
So the category of left modules $_A\mathcal V \simeq Fun^\mathcal V(BA, \mathcal V)$ is $\mathcal V$-enriched.
You can see your forgetful functor $F : _A\mathcal V \to \mathcal V$ as an $\mathcal V$-enriched functor, and so consider the monoid of endomorphisms $End(F)$ which is defined with the help of the end formula just cited.
So the Tannaka reconstruction theorem states that $End(F) \simeq A$ as monoids of $\mathcal V$ and it is proven by the use of the enriched Yoneda lemma, the fact that the forgetful functor is $\mathcal V$-representable by the following left $A$-module : $A$ and it's self left action and that the endomorphisms of left $A$-modules of $A$ is $A$.
You can find the material about enrichement in many books and articles that treat enrichement, as the book of Emily Riehl on "Categorical Homotopy Theory", the nLab sends you also to Max Kelly's "Basic Concepts of Enriched Category Theory". For the tannaka part I was not very familiar with that until now so I don't have references other than the given in the nlab page.