If $(\mathcal{C},\otimes, I)$ is merely a monoidal category, then $\mathsf{Mon}(\mathcal{C})$ need not be a monoidal category, and if it is it is likely not in any natural way. For instance, taking $\mathcal{C}=\mathrm{Fun}(\mathcal{A},\mathcal{A})$ for a category $\mathcal{A}$, we can define a monoidal structure on $\mathcal{C}$ via composition of functors. If you want the forgetful functor $\mathsf{Mon}(\mathcal{C})\to\mathcal{C}$ to preserve a monoidal structure, you would be asking if, given two monads $T,U$ on $\mathcal{A}$, the composition $TU$ is again a monad. This question has been answered today here, and generally this is false.
If $(\mathcal{C},\otimes, I)$ is a symmetric monoidal category, then $\mathsf{Mon}(\mathcal{C})$ and $\mathsf{CMon}(\mathcal{C})$ are again symmetric monoidal categories, and the forgetful functor is monoidal. The tensor product of monoids is defined as follows: given two monoids $(M,m,e)$, $(N,n,f)$ in $\mathcal{C}$, then we can consider the object $M\otimes N\in\mathcal{C}$, with multiplication law
$$
(M\otimes N)\otimes (M\otimes N)\cong(M\otimes M)\otimes(N\otimes N)\xrightarrow{m\otimes n}M\otimes N,
$$
and unit
$$I\cong I\otimes I\xrightarrow{e\otimes f}M\otimes N.$$
You can check that $M\otimes N$ will be a monoid again, which is commutative if both $M$ and $N$ are. The trivial monoid on $I$ serves as unit object.
If $f\colon (M,m,e)\to(M',m',e')$ is a monoid homomorphism, then $f\otimes N\colon M\otimes N\to M'\otimes N$ will be as well, which is again checking a bunch of diagrams commute. Likewise, the associativity, commutativity and unitality isomorphisms of $\otimes$ in the category $\mathcal{C}$ also lift to monoid isomorphisms if we take the tensor product of monoids. In particular, the coherence axioms just follow from the fact that they hold for $\otimes$ in $\mathcal{C}$.
