Note: we need to assume that the tensor product is symmetric, so that we can commute hom objects past each other to define the tensor product of $\mathcal{V}$-categories.
Partial answer. I can prove that limits, tensor, and internal hom exist, but I'm finding colimits harder.
General stuff
Let $I$ be a small category. Let $\newcommand\V{\mathcal{V}}\V$ be a complete, cocomplete, closed monoidal category.
Let $D:I\to \V\mathbf{Cat}$ be a diagram of (I'm going to assume small) $\V$-enriched categories. There are both free and cofree $\V$-categories on a set of objects. The free category $F_S$ has objects $S$ and
morphism objects
$$F_S(s,t) =
\begin{cases}
1 & s=t \\
\varnothing & s\ne t,
\end{cases}
$$
where $\varnothing$ is the initial object. The composition law is uniquely defined,
and you can check that functors from this category only depend on where they send the set $S$.
For the cofree category, replace $\varnothing$ with $*$, the terminal object.
As a result, to construct the co/limit of $D$, we know that if they exist, their sets of objects will be the co/limit of the sets of objects of the categories in the diagram.
Constructing the limit
An object of
$\lim D$ will be a family $A=\{A_i\}_{i\in I}$ of objects $A_i\in D_i$ for each $i\in I$ such that
if $f:i\to j$ is a morphism in $I$, then $f_*A_i=A_j$. (I'll use pushforwards to denote application of a covariant functor I don't want to name to a morphism).
Given two such families $A$ and $B$, we need to construct the hom object
$(\lim D)(A,B)$. Moreover, $\lim D$ needs a $\V$ functor to every $D_i$, so
we need a map $(\lim D)(A,B)\to (D_i)(A_i,B_i)$ compatible with the maps
$f_* : D_i(A_i,B_i)\to D_j(A_j,B_j)$ induced by morphisms $f:i\to j$.
Therefore we should define $$(\lim D)(A,B) = \lim_i D_i(A_i,B_i).$$
We have a composition:
$$
\begin{align}
(\lim D)(B,C)\otimes (\lim D)(A,B)
&=
(\lim_i D_i(B_i,C_i))\otimes
(\lim_j D_j(A_j,B_j))
\\
&\to
\lim_i D_i(B_i,C_i)\otimes D_i(A_i,B_i)
\\
&\to
\lim_i D_i(A_i,C_i)
\\
&= (\lim D)(A,C).
\end{align}
$$
The first $\to$ comes from the universal property of the limit.
The second $\to$ is functoriality of the limit applied to all of the composition maps.
You can check that this is associative and unital.
We then need to check that all the projection maps are functorial, but this is essentially free from how we just defined composition, so I'll let you check that. The universal property is also easy to check from the definition. Just a check on what we used. We used that $V$ was complete and monoidal, cocomplete and closed didn't enter into it.
Partial progress on defining the colimit
Disclaimer:
I'm not sure if I've defined the hom objects correctly, since I can't put a composition together. Assuming $\V$ is symmetric monoidal might help. Also I think everything below works in a straightforward manner when the colimit is filtered. Anyway, these are my thoughts so far.
Side note (alternative strategy) We have that arbitrary coproducts exist by taking hom objects to be $\varnothing$ when the objects are from different categories, so we could also attack this problem by producing coequalizers, which might be a bit easier.
For colimits, it's a little harder, since it's hard to get an explicit handle on the objects in the colimit category. However, there should still be an analogous construction.
Let $A,B$ be equivalence classes of objects in the colimit of the sets of objects of the categories $D_i$.
Suppose $a_i,b_i\in D_i$ such that $a_i\in A$, and $b_i\in B$, (identifying objects with their images under the canonical inclusions). Then we expect a
morphism
$$D_i(a_i,b_i)\to (\newcommand\colim{\operatorname{colim}}\colim_I D)(A,B),$$
and it should be compatible with the corresponding morphism
$$D_j(f_*a_i,f_*b_i)\to (\colim D)(A,B).$$
In other words, to define $(\colim D)(A,B)$, we need to take a colimit,
but not over $I$ this time. Instead the objects of the category will be pairs
$(a_i,b_i)$ with $a_i,b_i\in D_i$, and $a_i\in A$, $b_i\in B$, with morphisms
$(a_i,b_i) \to (a'_j,b'_j)$ being those morphisms
$f:i\to j$ in $I$ such that $f_*a_i=a'_j$ and $f_*b_i=b'_j$.
Now we need to define composition. We note that since $\V$ is closed symmetric monoidal, the tensor product preserves colimits in both variables. Thus
$$
\begin{align}
(\colim D)(B,C)
\otimes (\colim D)(A,B)
&=
(\colim_{b_i,c_i} D_i(b_i,c_i))
\otimes
(\colim_{a_j,b_j} D_j(a_j,b_j))
\\
&\simeq
\colim_{b_i,c_i}
\colim_{a_j,b_j}
D_i(b_i,c_i)\otimes D_j(a_j,b_j)
\\
\end{align}
$$
The tensor product of $\V$-categories
If $\newcommand\A{\mathcal{A}}\newcommand\B{\mathcal{B}}\A$ and $\B$ are $\V$-categories, then we define
the objects of $\A\otimes\B$ to be pairs $(a,b)$ of objects with $a\in\A$, $b\in\B$, and define
$$(\A\otimes\B)((a,b),(a',b')) = \A(a,a')\otimes \B(b,b').$$
Assuming the monoidal product is symmetric, this forms a $\V$-category with composition given by
$$
\begin{align}
(\A\otimes\B)((a',b'),(a'',b''))&
\otimes
(\A\otimes\B)((a,b),(a',b'))
\\
&=
\A(a',a'')
\otimes
\B(b',b'')
\otimes
\A(a,a')
\otimes
\B(b,b')
\\
&\simeq
\A(a',a'')
\otimes
\A(a,a')
\otimes
\B(b',b'')
\otimes
\B(b,b')
\\
&\to
\A(a,a'')
\otimes
\B(b,b'')
\\
&=
(\A\otimes \B)((a,b),(a'',b'')).
\end{align}
$$
The internal hom of $\V$-categories
First, let $1$ be the free $\V$-category on a single object. Then
$1$ represents the set of objects functor.
Therefore, if the internal hom exists, since
$$\V-\newcommand\Cat{\mathbf{Cat}}\Cat(1,[\A,\B])
\simeq \V-\Cat(1\otimes \A,\B)\simeq \V-\Cat(\A,\B),$$
we have that $[\A,\B]$ has the set of $\V$-functors from $\A$ to $\B$ as its underlying set of objects. If $F,G:\A\to \B$ are $\V$-functors,
we would expect by analogy with $\mathbf{Set}$-enriched categories that the
natural transformations hom object should be computed by
$$
[\A,\B](F,G)
=
\int_{a\in\A} \B(F(a),G(a)).
$$
Just to be clear about what this end means, since $F$ and $G$ are $\V$-enriched, and not ordinary functors, it means that it's the universal object $E$
with maps $\pi_a : E\to \B(F(a),G(a))$ such that for all pairs of objects $a,a'\in \A$,
the diagram
$$
\require{AMScd}
\begin{CD}
E \otimes \A(a,a') @> \pi_a\otimes G >> \B(F(a),G(a))\otimes \B(G(a),G(a'))
\\
@V\pi_a'\otimes F VV @VV\circ V
\\
\B(F(a'),G(a'))\otimes \B(F(a),F(a'))
@>\circ >> \B(F(a),G(a'))
\\
\end{CD}
$$
commutes.
You can check that identities/composition in $\B$ induce identities/composition on these hom objects. Thus we get a $\V$-category. We'll prove that this $\V$-category is in fact the internal hom.
We're just left with showing that this is right adjoint to
the tensor product.
Let $F:\newcommand\C{\mathcal{C}}\A\otimes\B\to \C$.
$F$ is a function $F$ from pairs of objects in $\A$ and $\B$ to objects in
$\C$ and a family of maps
$$
F_{(a,b),(a',b')} : (\A\otimes B)((a,b),(a',b'))
\to \C(F(a,b),F(a',b')).
$$
We need to define a $\V$-functor $G:\A\to [\B,\C]$.
For a fixed object $a$, the functor
$F(a,-)$ is a $\V$-functor from $\B$ to $\C$,
with hom morphism for $b$, $b'$ given by
$$\B(b,b')\simeq 1\otimes \B(b,b')
\to \A(a,a)\otimes \B(b,b')\xrightarrow{F} \C(F(a,b),F(a,b')).$$
So on objects we have $G(a)=F(a,-)$. We define the map on morphisms using the universal property of the end.
We need to produce a map $p_b$ for each object $b\in B$
$\A(a,a')\to \C(F(a,b),F(a',b)),$
but these are just the hom maps of $F(-,b)$, and
the commutativity condition becomes that
$$
\begin{CD}
\A(a,a') \otimes \B(b,b') @> F(-,b)\otimes F(a',-) >>
\C(F(a,b),F(a',b))\otimes \C(F(a',b),F(a',b'))
\\
@V F(-,b')\otimes F(a,-) VV @VV\circ V
\\
\C(F(a,b'),F(a',b'))\otimes \C(F(a,b),F(a,b'))
@>\circ >> \C(F(a,b),F(a',b'))
\\
\end{CD}
$$
commutes, which follows from functoriality of $F$.
Thus we get hom maps for $G$ from the universal property, and you can check that they make $G$ into a functor.
For the other direction, this argument can essentially be traced backwards to recover $F$ from $G$, and then you could check naturality.