Is a monoidal operation isomorphic to product under these conditions? Let $C$ be a monoidal category with monoidal operation $\otimes$. Suppose $T = A \otimes - : C \rightarrow C$ forms a comonad. That is, suppose there are natural transformations $\Delta : T \implies T \circ T$ and $\eta: T \rightarrow \text{Id}$ forming a comonad. 
What else must we require for the monoidal operation to be product, $\Delta$ to be the diagonal map, and $\eta$ to be projection onto a single coordinate?
 A: Thought this was an interesting question, so I thought I'd expand on Arnaud D.'s comment with a proof. It might not answer the asked question, depending on the motivation of the asker, but it's interesting enough on its own.
Note The version of the result that I found on nLab doesn't require uniqueness of the comonoid structure, instead it requires a symmetric monoidal category and naturality of the comonoid structures.
Note on notation
I'll omit the unitors/associators for notational convenience. 
Comonoid structures on $A$ give rise to comonad structures on $A\otimes -$
Hopefully it is clear that if $A$ has a comonoid structure, 
$\Delta : A\to A\otimes A$, $\eta : A \to 1$, then $\Delta \otimes -$ and $\eta \otimes -$ give a comonad structure on $T=A\otimes -$. 
Conversely comonad structures on $A\otimes -$ give rise to comonoid structures on $A$
If $T=A\otimes -$, then $\Delta_1 : T1 \to TT1$ is a map $A\to A\otimes A$, and similarly for $\eta_1 : A\to 1$. The comonad axioms imply that these satisfy the comonoid axioms.
At the moment, it's not clear to me whether or not the comonoid structure determines the comonad structure, but that might be because it's late and my brain has stopped functioning.
Every object has a unique comonad structure for the cartesian monoidal structure, $(\times,*)$
We already know existence, the comonad structure is given by the diagonal map $\Delta = (1_A,1_A)$ and $\eta : A\to *$ the unique map to the terminal object. 
Uniqueness follows from the fact that $\eta : A\to *$ is forced to be the unique map to the terminal object, so the unit axioms $(1_A\times \eta) \circ \Delta = 1_A = (\eta \times 1_A)\circ\Delta$ force $\Delta$ to have components $1_A$ and $1_A$, and therefore be the diagonal map.
It's not hard to check that $\Delta_A : A\to A\times A$ and $\eta_A : A\to *$ are natural in $A$. You can also check that these are monoidal. Additionally the product is always symmetric monoidal.
Conversely, if every object has a natural comonoid structure for $(\otimes,1)$,
then $(\otimes,1)\simeq (\times,*)$.
Precisely: If $(\mathcal{C}, \otimes, 1)$ is symmetric monoidal, and there are monoidal natural transformations $\Delta : \operatorname{Id} \to (-\otimes -)$ and 
$\eta : \operatorname{Id} \to 1$ that define a comonoid structure on every object, then $1$ is a terminal object, and $A\otimes B$ has natural projection maps making it a product.
For each object $A$, let $\Delta_A : A\to A\otimes A$ and $\eta_A : A\to 1$ be the natural comonoid structure maps.
First, monoidality of the natural transformations forces $\eta_1 = 1_1$ is the identity, and $\Delta_1 : 1\newcommand\toby{\xrightarrow}\toby{\sim} 1\otimes 1$ is the unit isomorphism.
Now for any $\alpha : A\to 1$, naturality forces the following square to commute:
$$\require{AMScd}
\begin{CD}
A @>\eta_A>> 1 \\
@V\alpha VV @VV 1_1V \\
1 @>\eta_1=1_1>> 1 \\
\end{CD}
$$
Therefore $\alpha = \eta_1$. Thus $1$ is terminal.
We also get maps $1_A\otimes \eta_B: A\otimes B \to A$ and $\eta_A\otimes 1_B : A\otimes B\to B$. We need to show that these are projection maps satisfying the universal property making $A\otimes B$ a product of $A$ and $B$.
Suppose we have $f: X\to A$ and $g:X\to B$, then we should have that the unique map from the universal property is $(f\otimes g)\Delta_X : X\to A\otimes B$.
Composing our projection to $A$, we get
$$(1_A\otimes \eta_B)(f\otimes g)\Delta_X = (f\otimes 1_1)(1_A\otimes \eta_B)\Delta_X=f1_X=f,$$
with the first equality by functoriality of $\otimes$ and $\eta_B\circ g = \eta_B$ by uniqueness of the map to the terminal object, and the second equality by the unit axiom for the comonoid structure.
Symmetrically, when we compose with the other projection, we'll get $g$.
All that's left is uniqueness.
For uniqueness, let $\alpha : X\to A\otimes B$ be another morphism with the given property, then you can check that this means that 
$$(1_A\otimes \eta_B\otimes \eta_A\otimes 1_B)(\alpha\otimes \alpha) = (f\otimes g),$$
and what we want is to show that $\alpha = (f\otimes g)\Delta_X$, so it suffices to show 
that $$\alpha = (1_A\otimes \eta_B\otimes \eta_A\otimes 1_B)(\alpha\otimes \alpha)\Delta_X.$$
For this we consider the following diagram:
$$
\begin{CD}
X @>\Delta_X >> X\otimes X
\\
@V\alpha VV @VV\alpha\otimes\alpha V
\\
A\otimes B @>\Delta_{A\otimes B}>> A\otimes B\otimes A\otimes B
\\
@V 1_{A\otimes B} VV
@VV 1_A\otimes \tau\otimes 1_B V
\\
A\otimes B @>\Delta_A\otimes \Delta_B >> A\otimes A\otimes B\otimes B
\\
@V1_{A\otimes B}VV
@VV(1_A\otimes \eta_A)\otimes(\eta_B\otimes 1_B)V
\\
A\otimes B@>>1_{A\otimes B}> A\otimes B
\end{CD}
$$
The top square commutes by naturality of $\Delta$, the middle square is monoidality of $\Delta$, and the bottom square is the unit axiom.
The outside rectangle is the identity we wanted to prove to prove uniqueness.
Therefore $A\otimes B$ is a product of $A$ and $B$. $\blacksquare$
End note
I can't see a proof that having a unique comonoid structure on every object forces the structures to be natural and monoidal, but this might be true.
