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.)