After some research, I think it has not been observed until now. However, all of the bricks needed to make the argument are almost ready.
In paper "Monoidal bicategories and Hopf algebroids" Brain Day and Ross Street defined a notion of convolution in the context of Gray monoids. For a reason that shall become clear later, I am willing to call it "virtual convolution". Here is the definition. Let $\langle A, \delta \colon A \rightarrow A \otimes A, \epsilon \colon A \rightarrow I \rangle$ be a weak comonoid, and $\langle B, \mu \colon B \otimes B \rightarrow B, \eta \colon I \rightarrow B \rangle$ be a weak monoid in a monoidal bi-category with tensor $\otimes$ and unit $I$, then $\langle \hom(A, B), \star, i \rangle$ is a monoidal category by:
\begin{array}{ccc}
f\star g &=& \mu \circ (f \otimes g) \circ \delta \newline
i &=& \eta \circ \epsilon
\end{array}
So the "convolution structure" exists only virtually --- on $\hom$-categories. If the monoidal bi-category admits all right Kan liftings, then such induced monoidal category $\langle \hom(I, B), \star, i \rangle$ for trivial comonoid on $I$ is monoidal (bi)closed by:
$$f \overset{L}\multimap h = \mathit{Rift}_{\mu \circ (f \otimes \mathit{id})}(h)$$
$$f \overset{R}\multimap h = \mathit{Rift}_{\mu \circ (\mathit{id} \otimes f)}(h)$$
Taking for the monoidal bi-category the bi-category of profunctors, we obtain the well-known formula for convolution. However, in the general setting, such induced structure is far weaker than one would wish to have --- for example in the category of profunctors enriched over a monoidal category $\mathbb{V}$ the induced convolution instead of giving a monoidal structure on the category of enriched presheaves:
$$\mathbb{V}^{B^{op}}$$
merely gives a monoidal structure on the underlying category:
$$\hom(I, \mathbb{V}^{B^{op}})$$
Actually, there is a work-around for this issue in the context of enriched categories, as suggested in the paper, but the general weakness of "virtual convolution" is obvious.
The solution is to find a way to "materialize" the convolution. I shall sketch the idea for internal categories. I think all of the following works for split fibrations and split structures, so let me replace the codomain fibration $\mathbb{C}^\rightarrow \rightarrow \mathbb{C}$ from the question by its split version corresponding to the internal "family functor":
$$\mathit{fam}(\mathbb{C}) \colon \mathbb{C}^{op} \rightarrow \mathbf{Cat}$$
Likewise, for a category $A$ internal to $\mathbb{C}$ I shall write:
$$\mathit{fam}(A) \colon \mathbb{C}^{op} \rightarrow \mathbf{Cat}$$
for the functor corresponding to the externalisation of $A$. We want to show that given a promonoidal structure $$\langle A, \mu \colon A \times A \nrightarrow A, \eta \colon 1 \nrightarrow A \rangle$$
there is a corresponding monoidal closed structure on:
$$\mathit{fam}(\mathbb{C})^{\mathit{fam}(A)^{op}}$$
which just means, that each fibre of $\mathit{fam}(\mathbb{C})^{\mathit{fam}(A)^{op}}$ is a monoidal closed category and reindexing functors preserve these monoidal structures. By fibred Yoneda lemma, for $K \in \mathbb{C}$:
$$\mathit{fam}(\mathbb{C})^{\mathit{fam}(A)^{op}}(K) = \mathit{Prof}(K, A)$$
where $K$ is interpreted as a discrete internal category. There is a correspondence:
$$\mathit{Prof}(K, A) \approx \mathit{Prof}(1, K^{op} \times A) = \mathit{Prof}(1, K \times A)$$
where the last equality holds because $K^{op} = K$ for any discrete category $K$. Since $K$ has a trivial promonoidal structure:
$$K \times K \overset{\Delta^*}\nrightarrow K$$
we obtain a "product" promonoidal structure on $K \times A$:
\begin{array}{rcc}
K \times A \times K \times A &\overset{\Delta^* \times \mu}\nrightarrow& K \times A \newline
1 &\overset{\langle !^*, \eta \rangle}\nrightarrow& K \times A
\end{array}
In more details, since $\mathbb{C}$ is cartesian, every object $K \in \mathbb{C}$ carries a unique comonoid structure:
\begin{array}{l}
K \overset{\Delta}\rightarrow K \times K \newline
K \overset{!}\rightarrow 1
\end{array}
which has a promonoidal right adjoint structure $\langle \Delta^\*, !^\* \rangle$ in the (bi)category of internal profunctors. The product of the above two promonoidal structures is given by the usual cartesian product of internal categories (note, it is not a product in the bicategory of internal profunctors) followed by the internal product functor $\mathit{fam}(\mathbb{C}) \times \mathit{fam}(\mathbb{C}) \overset{\mathit{prod}}\rightarrow \mathit{fam}(\mathbb{C})$.
Then by "virtual convolution" there is a monoidal (bi)closed structure on $\mathit{Prof}(1, K^{op} \times A)$. Therefore each fibre $\mathit{fam}(\mathbb{C})^{\mathit{fam}(A)^{op}}(K)$ is a monoidal (bi)closed category. It is easy to check that reindexing functors preserve these structures.
Let me work out the concept of internal Day convolution in case $\mathbb{C} = \mathbf{Set}$ and a promonoidal structure on a small category is monoidal. The split family fibration (or more accurately, the indexed functor corresponding to the family fibration) for a locally small category $A$:
$$\mathit{fam}(A) \colon \mathbf{Set}^{op} \rightarrow \mathbf{Cat}$$
is defined as follows:
\begin{array}{rcl}
\mathit{fam}(A)(K \in \mathbf{Set}) &=& A^K \newline
\mathit{fam}(A)(K \overset{f}\rightarrow L) &=& A^L \overset{(-) \circ f}\rightarrow A^K\newline
\end{array}
where $K, L$ are sets and $K \overset{f}\rightarrow L$ is a function between sets. One may think of category $A^K$ as of the category of $K$-indexed tuples of objects and morphisms from A. Now, given any monoidal structure on a small category $$\langle A, \otimes \colon A \times A \rightarrow A, I \colon 1 \rightarrow A \rangle$$
the usual notion of convolution induces a monoidal structure on $\mathbf{Set}^{A^{op}}$:
$$\langle F, G \rangle \mapsto F \otimes G = \int^{B, C \in A} F(B) \times G(C) \times \hom(-, B \otimes C)$$
The split fibration:
$$\mathit{fam}(\mathbf{Set})^{\mathit{fam}(A)^{op}} \colon \mathbf{Set}^{op} \rightarrow \mathbf{Cat}$$
may be characterised as follows:
\begin{array}{rcl}
\mathit{fam}(\mathbf{Set})^{\mathit{fam}(A)^{op}}(K \in \mathbf{Set}) &=& \mathbf{Set}^{A^{op} \times K} \newline
\mathit{fam}(\mathbf{Set})^{\mathit{fam}(A)^{op}}(K \overset{f}\rightarrow L) &=& \mathbf{Set}^{A^{op} \times L} \overset{(-) \circ (\mathit{id} \times f)}\rightarrow \mathbf{Set}^{A^{op} \times K}\newline
\end{array}
Since $\mathbf{Set}^{A^{op} \times K} \approx (\mathbf{Set}^{A^{op}})^K$ we may think of $\mathbf{Set}^{A^{op} \times K}$ as of $K$-indexed tuples of functors ${A^{op} \rightarrow \mathbf{Set}}$. In fact:
$$\mathit{fam}(\mathbf{Set})^{\mathit{fam}(A)^{op}} \approx \mathit{fam}(\mathbf{Set}^{A^{op}})$$
It is natural then to extend the monoidal structure induced on $\mathbf{Set}^{A^{op}}$ pointwise to $(\mathbf{Set}^{A^{op}})^K$:
$$(F \otimes G)(k) = \int^{B, C \in A} F(k)(B) \times G(k)(C) \times \hom(-, B \otimes C)$$
where $k \in K$.
On the other hand, using the internal formula for convolution, we get (up to a permutation of arguments):
\begin{array}{c}
\int^{B, C \in A, \beta, \gamma \in K} F(B, \beta) \times G(C, \gamma) \times \hom(\Delta(k), \langle \beta, \gamma \rangle) \times \hom(-, B \otimes C) \newline\hline\newline\hline
\int^{B, C \in A, \beta, \gamma \in K} F(B, \beta) \times G(C, \gamma) \times \hom(k, \beta) \times \hom(k, \gamma) \times \hom(-, B \otimes C) \newline\hline\newline\hline
\int^{B, C \in A} F(B, k) \times G(C, k) \times \hom(-, B \otimes C) \newline
\end{array}
where the first equivalence is the definition of diagonal $\Delta$ --- recall that the diagonal $\Delta(k) = \langle k, k \rangle$ is represented by profunctor $\hom(\langle \overset{1}-, \overset{2}-\rangle, \Delta(\overset{3}-))$, which has profunctorial right adjoint $\hom(\Delta(\overset{1}-), \langle \overset{2}-, \overset{3}-\rangle) \approx \hom(\overset{1}-, \overset{2}-) \times \hom(\overset{1}-, \overset{3}-)$ --- and the second one is by "Yoneda reduction" applied twice.
Final remarks:
Seeing the above proof, one may wonder where the assumptions about the category $\mathbb{C}$ from the question were actually used:
local cartesian closedness guaranteed existence of all right Kan liftings in the bi-category of internal profunctors; without this assumption, the induced monoidal structure on $\mathit{fam}(\mathbb{C})^{\mathit{fam}(A)^{op}}$ would be generally non-closed; to see that local cartesian closedness is really crucial here, recall that fibration $\mathit{fam}(\mathbb{C})$ is a cartesian closed fibration iff $\mathbb{C}$ is locally cartesian closed ---- this means that without local cartesian closedness even trivial convolution of the monoidal structure on the terminal category is not closed; moreover, which has not been stated in the answer, local cartesian closedness made it possible to speak about internal Yoneda embedding
finite colimits (coequalisers) allowed us to define compositions of internal profunctors
To really obtain a split monoidal closed structure via convolution without moving through the equivalence between Gray monoids and monoidal bi-categories ("Coherence for Tricategories", Gordon, Power, Street), one has (of course!) to replace the monoidal bi-category of internal profunctors by equivalent Gray monoid consisting of internal categories of presheaves and internally cocontinous functors.
I think that the right setting for the concept of Day convolution is a "Yoneda monoidal bi-triangle" as sketched in this answer.
Best Answer
Regarding Q2: probably there is a way to avoid going deep into coherence conditions: instead of proving by hand the equivalence between promonoidal structures on $C$ and biclosed monoidal structures on $\hat C$, one can resort to a more conceptual pov.
What happens for pro/monoidal categories is that there is a pseudomonad $S$ on $\sf Cat$ with the property that $S$ lifts to a pseudomonad $\hat S$ on $\sf Prof$ (the Kleisli bicategory of $P=\hat{(-)} = [(-)^{op},{\sf Set}]$), and pseudo-$S$-algebra structures correspond to pseudo-$\hat S$-algebra structures (this is an equivalence of categories, in the appropriate sense; see here).
I believe a similar argument holds for every (almost every?) monad $S$ equipped with a distributive law over $P$ (the presheaf construction); this does not fall short from an equivalence $$ \{S\text{-algebra structures on } PX\} \cong \{\hat S\text{-algebra structures on } X\} $$ where $PX$ is regarded as an object of $\sf Cat$, and $X$ as an object of ${\sf Kl}(P)$.
Regarding Q1: have you tried to find the distributive and annullator morphisms for the putative 2-rig structure on $\widehat{C}$?
I was trying to find at least one distributive morphism, and I have no idea how to reduce $F\hat{\otimes}(H\hat{\oplus} K)$ to/from $F\hat{\otimes} H \,\hat{\oplus}\, F\hat\otimes K$, if $F,H,K : \widehat{C}$. If I'm not wrong (this is very back-of-the-envelope coend calculus), $$\begin{align*} F\hat\otimes H &= \int^{UA}FU\times HA\times [\_, U\otimes A]\\ F\hat\otimes K &= \int^{U'B}FU'\times KB\times [\_, U'\otimes B] \end{align*}$$ whereas $$\begin{align*} F \hat\otimes \,(H\hat\oplus K) &= \int^{UV} FU \times (H\hat\oplus K)V \times [\_, U\otimes V] \\ &=\int^{UVAB} FU \times HA \times KB \times [V, A\oplus B] \times [\_, U\otimes V] \\ &=\int^{UAB} FU \times HA \times KB \times [\_, U\otimes (A\oplus B)] \\ &=\int^{UAB} FU \times HA \times KB \times [\_, U\otimes A \oplus U\otimes B] \\ \end{align*}$$ ...and now we're stuck, unless we have either
Actually, you need both in order for the computation to proceed; but the conjunction of R1 and R2 is quite strong, as you can see.
Edit: the situation with annullators (for Laplaza, morphisms ${\bf 0}\otimes X \to {\bf 0}$ and $X\otimes {\bf 0} \to \bf 0$) is even worse!
Let's open $F \hat\otimes {\bf 0}$ recalling that in this case $\bf 0$ is the representable $y{\bf 0}$ on the additive unit of $C$: $$\begin{align*} \int^{UV} FU \times [V,{\bf 0}] \times [\_,U\otimes V] &=\int^U FU \times [\_, U\otimes {\bf 0}] \\ &\overset{\rho_U}\to\int^U FU \times [\_, {\bf 0}]\\ &=\varinjlim F \times [\_, {\bf 0}] \end{align*}$$ the cartesian structure on $\sf Set$ now entails that this is $\bf 0$ if and only if either factor is empty, but I see no way in which this can be or even map into $y{\bf 0}$ again, as it should.