In checking the details of the correspondence between operads over a symmetric monoidal category and monads on some associated endofunctor of the category, I cannot make the obvious proof work without assuming that the monoidal product distributes over coproducts. But no such assumption is mentioned in my sources (for example Operads, Algebras, Modules by May (PDF).) Am I missing something in my argument?
Let $\mathcal{C}$ (mathcal C) be an operad (for simplicity take it non-symmetric) over a symmetric monoidal category $\mathcal{V},$ with composition $\gamma\colon \mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes\dotsb\otimes\mathcal{C}(m_n)\to\mathcal{C}(m_1+\dotsb+m_n).$ We define a functor $C\colon \mathcal{V}\to\mathcal{V}$ (Roman C) by $CX = \coprod_i \mathcal{C}(i)\otimes X^{\otimes i}$.
Then one wants to verify that the operad structure $\gamma$ gives a monad on $C$. That is, we need a natural morphism $C^2X\to CX,$ or $\coprod_i \mathcal{C}(i)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes i}\to \coprod_k \mathcal{C}(k)\otimes X^{\otimes k}$. By universal prop of coproducts, it will suffice to exhibit an arrow $\mathcal{C}(n)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}\to \coprod_k \mathcal{C}(k)\otimes X^{\otimes k}$ for all $n$.
Clearly we have $\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n}\to \mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes\dotsb\otimes \mathcal{C}(m_n) X^{\otimes m_1+\dotsb+m_n}\to \\ \mathcal{C}(m_1+\dotsb+m_n)\otimes X^{\otimes m_1+\dotsb+m_n}\to\coprod_k\mathcal{C}(k)\otimes X^{\otimes k},$ where the first arrow is by symmetry of the monoidal structure, the second arrow is the operad composition $\gamma,$ and the third arrow is the canonical inclusion into the coproduct.
Therefore by universal prop of coproducts, we have $\coprod_{m_1,\dotsc,m_n}\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n}\to\coprod_k\mathcal{C}(k)\otimes X^{\otimes k}.$
In general, again using inclusion morphisms of coproducts, we have arrows $\mathcal{C}(m_\ell)\otimes X^{\otimes m_\ell}\to \coprod_j \mathcal{C}(j)\otimes X^{\otimes j}.$ Then by functorality of the monoidal product, we have $\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n}\to \mathcal{C}(n)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}$. By universal property of coproducts, we therefore have an arrow from the coproduct $\coprod_{m_1,\dotsc,m_n}\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n}\to \mathcal{C}(n)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}$.
To summarize, we have the obvious maps $\mathcal{C}(n)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}\leftarrow \coprod_{m_1,\dotsc,m_n}\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n} \to \\ \coprod_k \mathcal{C}(k)\otimes X^{\otimes k}.$ Unless we know that the arrow on the left is an isomorphism, we do not get the structure map for a monad on $C$. And that arrow on the left will generally not be an isomorphism if the monoidal product in $\mathcal{V}$ does not distribute over the coproduct. For example, if the monoidal product is the coproduct itself.