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 :
- The image of the identity of $V$ under the natural isomorphism $\eta^V_{1,V} : Hom_{\mathcal V}(V,V) \to Hom_{\mathcal V}(1,\{V,V\})$ gives the enriched identity of $V$.
- The isomorphism $\eta^1_{X,V}$ gives that $\{1,V\}$ and $V$ represent the same functor, so they are ismorphic in $\mathcal V$.
- The image of the identity of $\{V,W\}$ under the natural isomorphism $(\eta^V_{\{V,W\},W})^{-1} : Hom_{\mathcal V} (\{V,W\},\{V,W\}) \to Hom_{\mathcal V} (\{V,W\}\otimes V, W)$ gives the evaluation map $ev_{V,W}$.
- This last map will help build a map $\{V,W\}\otimes \{U,V\} \otimes U \to W$ by the forumla $ev_{V,W} \circ (id_{\{V,W\}} \otimes ev_{U,V})$ which corresponds under $\eta$ to a morphism $c_{U,V,W}:\{V,W\}\otimes \{U,V\} \to \{U,W\}$ which is the enriched composition.
I will let you check that there is associativity and unitality of this as morphisms in $\mathcal V$.
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.
Best Answer
To define the product functor we need to fix a binary product, together with the projections, for each pair of objects.
The first thing to verify here is that it indeed defines a functor, i.e. uniquely determines the image of any pair of morphisms (in a functorial way).
I guess, the most problematic parts of the statement are the coherence conditions.
To get the proof smoother, let's also fix ternary and quarternary products (and nullary product as fixing the terminal object $1$, and unary products with the identity as projection).
For example, if we write $(a\times b\times c)\times d$, that is a binary product whose first term is a ternary product, and by the universal property we can easily see that it has a unique isomorphism from the quarternary product $a\times b\times c\times d$ that commutes with our fixed projections.
Similarly, for any act of 'putting in a pair of parenthesis' into a product of objects there belongs a unique isomorphism.
Now we define the associators $\alpha_{a,b,c}$ as the composite $(a\times b)\times c\longleftarrow a\times b\times c\longrightarrow a\times (b\times c)$ of the above isomorphisms.
Once again using the universal property of products, one can prove that if we put two pairs of parentheses in a product through the above isomorphisms, then the order doesn't matter: we'll receive commutative squares (like in the picture below).
To conclude the coherence pentagon condition, I found this picture among my old notes (on Leinster's unbiased bicategories), where the product sign $\times$ is omitted (so for example the top arrow is read as $((a\times b)\times c)\times d\overset{\alpha_{a,b,c}\times d}\longrightarrow (a\times(b\times c))\times d$).
The coherence conditions for unit are similar but easier, they might be thought as the case when we put an empty pair of parentheses into a product, like in "$ab\to ab()$" which would mean $a\times b\longrightarrow a\times b\times 1$.