This might not be the right approach, but here's my take on the question:

If $(\mathcal{C}, \otimes, I, \dots)$ is a (strict, for the sake of readability) monoidal category, we can endow $[\mathcal{C}, \mathcal{C}]$ with two different monoidal structures:

 1. The "usual" monoidal structure, where $\circ$ is the product and $\mathbf{1}_\mathcal{C}$ is the unit. This is the monoidal category whose monoids are called monads. Notice how this doesn't actually use the fact that $\mathcal{C}$ is endowed with a monoidal structure.

 2. The "other" (I sometimes refer to it as "pointwise") monoidal structure has "pointwise $\otimes$" (let's call it $\boxtimes$) as product, and $\Delta_I$ (the constant functor on the unit object of $\mathcal{C}$) as monoidal units. In this category, monoids are what Haskell programmers call "alternative" functors, but with one additional property. This can be further generalized to the case of functors $\mathcal{B} \to \mathcal{C}$ where $\mathcal{B}$ is *not* required to be monoidal, and everything that follows should work just fine (but I won't venture there).

Let me elaborate on the latter, since the former is (probably) already well-known to most haskell programmers here (other than being significantly less relevant):

First we'll need to properly define $\boxtimes$: if we fix functors $\mathcal{F}, \mathcal{G} : \mathcal{C} \to \mathcal{C}$ we can define $(\mathcal{F} \boxtimes \mathcal{G}) (a) := \mathcal{F}(a) \otimes \mathcal{G}(a)$ (the definition for morphisms is analogous). This, together with $\Delta_I$ as a unit, defines a monoidal category: all axioms (and further properties, such as symmetry and strictness) follow trivially from the monoidal structure on $\mathcal{C}$. 

Now, monoids in this monoidal category are what haskell programmers call "alternative" functors, but that's not all: they are so in a functorial way, meaning that if $f : a \to b : \mathcal{C}$ and $\mathcal{M}$ is a monoid in $([\mathcal{C}, \mathcal{C}], \boxtimes, \Delta_I)$ then $\mathcal{M}(f)$ is a morphism of monoids.  To see this, note how [this diagram][1] needs to commute in a pointwise fashion, and the same applies to [this other diagram][2] too. We're in the strict case, but nothing significant changes in the general one (except that you need to keep track of the coherence laws).

A more abstract (but equivalent) formulation is that each object $\mathcal{M} : \mathbf{Mon}([\mathcal{C}, \mathcal{C}])$ naturally defines a functor $\hat{\mathcal{M}} : \mathcal{C} \to \mathbf{Mon}(\mathcal{C})$, and this point of view leads to the observation that this actually defines a functor $(\hat{-}) : \mathbf{Mon}([\mathcal{C}, \mathcal{C}]) \to [\mathcal{C}, \mathbf{Mon}(\mathcal{C})]$. 

What this amounts to is just that a "monoidal" natural transformation $\nu : \mathcal{M} \to \mathcal{N} : \mathbf{Mon}([\mathcal{C}, \mathcal{C}])$ between monoid object in this "pointwise" monoidal category of (endo)functors also defines a natural transformation $\hat{\nu} : \hat{\mathcal{M}} \to \hat{\mathcal{N}} : [\mathcal{C}, \mathbf{Mon}(\mathcal{C})]$: the argument is similar but you need to look at the diagrams for [morphisms of monoids][3].

Things are not over yet: it seems that $(\hat{-})$ is actually an equivalence of categories! You can convince yourself of this if you squint hard enough, and the argument is roughly "brute force":

Let's construct a strict inverse to $(\hat{-})$, call it $(\tilde{-})$: take $\mathcal{F} : \mathcal{C} \to \mathbf{Mon}(\mathcal{C})$, compose it with the forgetful functor $\mathbf{U} : \mathbf{Mon}(\mathcal{C}) \to \mathcal{C}$. We'll now use the "forgotten structure" on $\mathcal{F}$ to endow $\mathbf{U} \circ \mathcal{F}$ with the structure of a monoid in $[\mathcal{C}, \mathcal{C}]$: to do this, notice how the monoid structure on $\mathcal{F}(a)$ (for $a : \mathcal{C}$) needs to be such that each morphism $\mathcal{F}(f) : \mathcal{F}(a) \to \mathcal{F}(b)$ is a morphism of monoids: this amounts to saying that the correspondences $a \mapsto \mathbf{m}^{\mathcal{F}(a)}$ and $a \mapsto \mathbf{e}^{\mathcal{F}(a)}$ are natural in $a$ and hence natural transformations (its now trivial to check that these endow $\mathbf{U} \circ \mathcal{F}$ with the structure of a monoid in $[\mathcal{C}, \mathcal{C}]$).

We now need to check that $(\hat{-}) \circ (\tilde{-}) = \mathbf{1}_{[\mathcal{C}, \mathbf{Mon}(\mathcal{C})]}$ and $(\tilde{-} \circ \hat{-}) = \mathbf{1}_{\mathbf{Mon}([\mathcal{C}, \mathcal{C}])}$: they should both be long and trivial computations, and I'll (boldly) omit them.

  [1]: https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAB12BbOnACwDGjYAFkAvgAJOAIwgAPPF3hT2AERgMcdAPoBJEGNLpMufIRQBGclVqMWbTj35CGosQaMgM2PASIAma2p6ZlZEDjUNLT0VWQUsJTgVJ0FhcQ9jHzMiMgsbEPtwx15U13FY+UVlYuc090NM0z9LUjzguzCIlJc3DK8TX3NkQLbbUIduEp7ymUqE6snasvqbGCgAc3giUAAzACcILiQyEBwIJCsxwoi9vnPqBjppDQAFAezwvax1vhw+-cOSECp3OiEuBU6nEeXGkUDoIAeT1e72aIAYMB2fwaIABR0QwLOSAArO1xkV2DA0NgGAQKvFEskStIdsALGJtDVSr1Ec8GG8sqivj8sZ5ccdqITEABmHnIgXmNEYv6k66c5ms9mc6aSWb0haU6l+bFi6US0EAFllfJRCvRmIRV0h3CY-wOeJJIKQlrRSOt8rYQt+DohEy4LuNbouZq9KqdKXVbI5iy56TEFDEQA
  [2]: https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAB12BbOnACwDGjYAFkAvgAJOAIwgAPPF3hTuvQcPErZCrErgqe-IQ1FiQY0uky58hFGQCMVWoxZtOh9Sc0z5i5R5qxqbmliAY2HgERABM5M70zKyIHKpGGpK+OnoGQRmhVpG2saRO1IluKYHp3mZizjBQAObwRKAAZgBOEFxIZCA4EEgO5a7JqVxMWn66AWl80u3ADmIA+tVeIRYd3b2IcQNDiADM1Ax00jAMAArWUXYgnVhNfDggo0nu3EwFIF09w2ogyQpxcnyq31+-z2-WB+zOFyutyK0RSTxebw+lQmakWyzWG2CPnY2n8+g8P3qYiAA
  [3]: https://ncatlab.org/nlab/show/monoid+in+a+monoidal+category#morphism_of_monoids