Monoidal categories whose tensor has a left adjoint

If $\otimes : V \times V \to V$ has a left adjoint and $V$ has finite products then $\otimes$ preserves them in the sense that the natural map

$$(X \times Y) \otimes (Z \times W) \to (X \otimes Z) \times (Y \otimes W)$$

is an isomorphism. By a monoidal-categorical version of the Eckmann-Hilton argument it seems to me that this implies that $\otimes$ is the product. Explicitly, if we let $1_{\times}$ denote the terminal object and $1_{\otimes}$ denote the monoidal unit then we get isomorphisms

$$1_{\otimes} \cong 1_{\otimes} \otimes 1_{\otimes} \cong (1_{\otimes} \times 1_{\times}) \otimes (1_{\times} \times 1_{\otimes}) \cong (1_{\otimes} \otimes 1_{\times}) \times (1_{\times} \otimes 1_{\otimes}) \cong 1_{\times} \times 1_{\times} \cong 1_{\times}$$

so $1_{\otimes} \cong 1_{\times}$ (and this isomorphism is unique if it exists so we don't even need to worry all that much about naturality). Now we can drop the outrageous subscripts and just refer to $1$. This gives a natural isomorphism

$$X \otimes Y \cong (X \times 1) \otimes (1 \times Y) \cong (X \otimes 1) \times (1 \otimes Y) \cong X \times Y$$

for any $X, Y$. Actually I'm not sure if this argument shows that the associator and unitor of $\otimes$ match up with the associator and unitor of the product but I'd guess a more elaborate version of this argument does.

I don't know if it's possible that $V$ doesn't have finite products. (There was previously an argument here involving Day convolution but Tim has pointed out gaps in it in the comments.)

Just to clean up the $\epsilon$ of room left after Qiaochu's answer -- we can get rid of the extra hypotheses. I'll write $I$ for the monoidal unit and $1$ for the terminal object.

Assume that $(\ell,r) \dashv \otimes$. Then the natural isomorphisms $A \cong I \otimes A \cong A \otimes I$ give rise, by adjunction, to maps $\ell A \to I$ and $r A \to I$, natural in $A$. We also have a unit map $A \to (\ell A) \otimes (r A)$, natural in $A$. Tensoring and composing, we get a map $A \to (\ell A) \otimes (r A) \to I \otimes I \cong I$, natural in $A$. That is, we have a cocone (with vertex $I$) on the identity functor for $V$. It follows that in the idempotent completion $\tilde V$ of $V$, there is a terminal object (which must be a retract of $I$).

Now, the idempotent completion $\tilde V$ again has a monoidal structure $\tilde \otimes$ with a left adjoint $(\tilde \ell, \tilde r)$. So the first part of Qiaochu's Eckmann-Hilton argument can be run in $\tilde V$: $I = I \otimes I = (I \times 1) \otimes (1 \times I) = (I \otimes 1) \times (1 \otimes I) = 1 \times 1 = 1$ (in the third expression, the products exist trivially, and in the fourth the product exists because $\otimes$ preserves products). That is, we must have $I_{\tilde V} = 1_{\tilde V}$. But $I_{\tilde V}$ is the image of $I_V$ in $\tilde V$, and the inclusion into the idempotent completion reflects terminal objects. Therefore $V$ has a terminal object, and $1_V = I_V$.

Then, as observed in the comments above, the second part of Qiaochu's Eckmann-Hilton argument can be run in $V$: $A \otimes B = (A \times 1) \otimes (1 \times B) = (A \otimes 1) \times (1 \otimes B) = A \times B$ (in the second expression, the products exist trivially, and in the third the product exists because $\otimes$ preserves products). That is, binary products exist in $V$ and agree with $\otimes$. In fact, the identity functor is an oplax monoidal functor from $(V,\otimes)$ to $(V,\times)$, which the argument shows is actually strong monoidal. Thus $(V,\otimes) \simeq (V,\times)$ as monoidal categories.