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.