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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. We can run the same argument above in the category $\widehat{V}$ of (small) presheaves on $V$ using Day convolution $\widehat{\otimes}$ with respect to $\otimes$, and we'll get a natural isomorphism $X \widehat{\otimes} Y \cong X \times Y$ in the presheaf category. The Yoneda embedding preserves and reflects limits so I think this implies that $\otimes$ is the product and in particular finite products exist in $V$.

Tim points out in the comments that it's not clear a priori that $\widehat{\otimes}$ still preserves products. I think we can argue as follows: the interchange map $(X \times Y) \widehat{\otimes} (Z \times W) \to (X \widehat{\otimes} Z) \times (Y \widehat{\otimes} Z)$ is a natural transformation between functors of four variables having values in $\widehat{V}$, is an isomorphism on $V$, the functors involved are cocontinuous in all four variables (Day convolution is closed monoidal), and presheaves are colimits of representables. So it's an isomorphism on $\widehat{V}$. (Admittedly I have not thought carefully about Day convolution using a monoidal category that isn't essentially small but I think if we restrict to small presheaves everything is fine; basically Day convolution just extends $\otimes$ to be cocontinuous in both variables, and small presheaves are the free cocompletion.)

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.)

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. We can run the same argument above in the category $\widehat{V}$ of presheaves on $V$ using Day convolution $\widehat{\otimes}$ with respect to $\otimes$, and we'll get a natural isomorphism $X \widehat{\otimes} Y \cong X \times Y$ in the presheaf category. The Yoneda embedding preserves and reflects limits so I think this implies that $\otimes$ is the product and in particular finite products exist in $V$.

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. We can run the same argument above in the category $\widehat{V}$ of (small) presheaves on $V$ using Day convolution $\widehat{\otimes}$ with respect to $\otimes$, and we'll get a natural isomorphism $X \widehat{\otimes} Y \cong X \times Y$ in the presheaf category. The Yoneda embedding preserves and reflects limits so I think this implies that $\otimes$ is the product and in particular finite products exist in $V$.

Tim points out in the comments that it's not clear a priori that $\widehat{\otimes}$ still preserves products. I think we can argue as follows: the interchange map $(X \times Y) \widehat{\otimes} (Z \times W) \to (X \widehat{\otimes} Z) \times (Y \widehat{\otimes} Z)$ is a natural transformation between functors of four variables having values in $\widehat{V}$, is an isomorphism on $V$, the functors involved are cocontinuous in all four variables (Day convolution is closed monoidal), and presheaves are colimits of representables. So it's an isomorphism on $\widehat{V}$. (Admittedly I have not thought carefully about Day convolution using a monoidal category that isn't essentially small but I think if we restrict to small presheaves everything is fine; basically Day convolution just extends $\otimes$ to be cocontinuous in both variables, and small presheaves are the free cocompletion.)

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.

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. We can run the same argument above in the category $\widehat{V}$ of presheaves on $V$ using Day convolution $\widehat{\otimes}$ with respect to $\otimes$, and we'll get a natural isomorphism $X \widehat{\otimes} Y \cong X \times Y$ in the presheaf category. The Yoneda embedding preserves and reflects limits so I think this implies that $\otimes$ is the product and in particular finite products exist in $V$.