You can find a result of that type as Proposition 3.46 in Kelly's "Basic concepts of enriched category theory" (available here). It's given for cotensors, but the dual result would be :
Proposition : If $\mathcal{V}$ is a monoidal category such that the functor $\mathcal{V}(I,\_)$ is conservative and each object of $\mathcal{V}$ has only a set of extremal epimorphic quotients, then a $\mathcal{V}$-enriched category $\mathcal{B}$ is tensored if its underlying category $\mathcal{B}_0$ is cocomplete.
In particular, this holds whenever $\mathcal{V}$ is monadic over $\mathbf{Set}$ and the unit object $I$ is the free object on one element, as in the case of $\mathcal{V}=\mathbf{Mod}_R$. Kelly also gives the category of Banach spaces with contractions as an example where $\mathcal{V}(I,\_)$ is conservative (see page 8).
The proof should go as follows.
In the strict monoidal category, we have that the corresponding compositions of $\alpha'$, $\alpha^{\prime -1}$, $l'$, $l^{\prime-1}$, $r'$, $r^{\prime -1}$ are all the identity.
Let $P_{Fi}$ be the corresponding parenthesization of $FX_1,\ldots,FX_n$ and $1'$s to $P_i$.
Monoidality of the equivalence gives natural isomorphisms $\alpha_i : FP_i \to P_{Fi}$ such that
$$
\require{AMScd}
\begin{CD}
FP_1 @>F f>> FP_2\\
@V\alpha_1 VV @V\alpha_2 VV \\
P_{F1} @> \mathrm{id} >> P_{F2}.
\end{CD}
$$
The same diagram commutes for $g$, so $Ff=Fg$.
Then by faithfulness of $F$, $f=g$.
Constructing $\alpha_i$
It sounds like you're having trouble with constructing the $\alpha_i$, so I'll expand on this a bit.
This goes inductively, so we just need to deal with the outermost layer of $P_1$ or $P_2$. We'll just look at $P_1$.
If $P_1 = X\otimes Y$, then $FP_1 = F(X\otimes Y)$ and $P_{F1} = X_F \otimes Y_F$, where the $F$ subscript refers to the appropriate product of $F$ applied to the atoms. Then we have the monoidal structure isomorphism $J : F(X\otimes Y) \to F(X)\otimes F(Y)$, and then we inductively construct $\alpha_X : F(X)\to X_F$ and $\alpha_Y : F(Y)\to Y_F$. The composite $(\alpha_X \otimes \alpha_Y) \circ J$ gives the desired isomorphism $FP_1\to P_{F1}$.
On the other hand, if $P_1 = I\otimes X$, then $P_{F1} = I'\otimes X_F$.
This time, if $\iota : F(I)\to I'$ is the natural isomorphism, we take the composite
$$F(I\otimes X)\overset{J}{\to} F(I)\otimes FX \overset{\iota\otimes \alpha_X}{\to} I'\otimes X_F= P_{F1}$$
For $P_1 = X\otimes I$ we do the same thing but modified for symmetry.
Hopefully this helps. I can expand on this if needed.
Edit, an expansion on why the diagram commutes.
This is also proved inductively.
Suppose the outermost function in $f$ is an associator, which I'll denote by $a$ and $a'$ in the two categories in the equivalence in $C_s$, $a'=\mathrm{id}$, since I used $\alpha$ for the natural isomorphisms.
Then $f=a\circ f_0$ and $f'=a'\circ f_0'$, with $f:P_1\to P_2$ and
$P_2 : X\otimes (Y \otimes Z$ for some products $X$, $Y$, and $Z$,
so $f_0 : P_1\to (X\otimes Y)\otimes Z$.
Then we have the commutative diagram
$$
\newcommand\id{\mathrm{id}}
\begin{CD}
FP_1 @>f_0>> F((X\otimes Y) \otimes Z) @>Fa>> F(X \otimes (Y\otimes Z)) \\
@VVV @VJVV @VJVV \\
@. F(X\otimes Y) \otimes FZ @. FX\otimes F(Y\otimes Z) \\
@V\alpha_1 VV @VJ\otimes \id VV @V\id \otimes J VV \\
@. (FX\otimes FY)\otimes FZ @>a'>> FX\otimes (FY\otimes FZ)\\
@VVV @V(\alpha_X\otimes \alpha_Y)\otimes \alpha_Z VV @V\alpha_X \otimes (\alpha_Y\otimes \alpha_Z)VV \\
P_{F1} @>f'_0>> (X_F\otimes Y_F)\otimes Z_F @>a'>> X_F \otimes (Y_F\otimes Z_F) \\
\end{CD}
$$
The left rectangle commutes by the inductive hypothesis applied to $f_0$, since the middle vertical composite is the definition of the $\alpha$ map for $(X\otimes Y)\otimes Z$. The top right rectangle commutes by the compatibility condition for $J$. The bottom right square commutes by naturality of the associator.
By symmetry, the same argument applies to $a^{-1}$.
Now we need to do the same thing for the left and right units and their inverses. By symmetry, it suffices to prove commutativity when the outermost map in $f$ is $l$.
Then $f = l\circ f_0$, with $f_0 : P_1\to I\otimes P_2$. This time, we get the diagram
$$
\begin{CD}
FP_1 @>f_0>> F(I\otimes P_2) @>Fl>> FP_2 \\
@VVV @VJVV @V\id VV \\
@. FI\otimes FP_2 @. FP_2 \\
@V\alpha_1 VV @V \iota \otimes \id VV @V\id VV \\
@. I'\otimes FP_2 @>l'>> FP_2 \\
@VVV @V\id \otimes \alpha_2 VV @V\alpha_2VV \\
P_{F1} @>f'_0>> I'\otimes P_{F2} @>l'>> P_{F2} \\
\end{CD}
$$
Again, the left rectangle commutes by the inductive hypothesis, the top right rectangle is the coherence condition for $\iota$, and the bottom right square is naturality of $l'$.
This completes the proof.
Best Answer
It's easiest to understand this question by relating $V$-categories to $V$-graphs. A $V$-graph $G$ is given by a set of objects $\mathrm{ob} G$ together with an object of $V$, denoted by $G(x,y)$, for every $x,y\in \mathrm{ob} G$.
As in the case of $V=\mathrm{Set}$, limits of $V$-categories are created by the forgetful functor into $V$-graphs. This means that the object set and the homs in a limit of $V$-categories are given by the corresponding limits of sets and of hom-objects in $V$, respectively. Coproducts of $V$-categories are just disjoint unions, so it's only coequalizers that really present some difficulties. That said, they really present some difficulties! See the following (freely available) paper of Wolff for the full construction, which proves along the way that $V$-categories are monadic over $V$-graphs. Wolff's paper
Frequently, the enriching category $V$ is in fact locally presentable, not just complete and cocomplete. In fact this is pretty much always the case unless $V=\mathrm{Top}$. In that case it may be shown that $V$-Cat is also locally presentable, which gives a higher-level proof that the latter is cocomplete. It was proved much later than Wolff's paper by Lack and Kelly that the following hold, if $V$ is cocomplete and tensors preserve colimits-for instance $V$ could be closed, but this is not required, and $V$ needn't be symmetric:
Neither of these papers produces a practically usable algorithm for computing colimits of $V$-categories, but this is unavoidable-for $V=$Set, considering $V$-categories with one object and focusing on coequalizers of maps between free objects, we have recovered the question of computing a monoid from generators and relations, which is known to be generally undecidable.