Be careful, when people say that a monad is just a monoid in the category of endofunctors over a given category they mean that a monad is a monoid object in the monoidal category of the endofunctors.
As you can see following the link above a monoid object is formed by an object $c$ of the monoidal category considered, with a binary operation (i.e. a morphism $c\otimes c \to c$) and a unit (i.e. a morphism $I \to c$, where $I$ is the unit of the monoidal structure over the category) satisfying the diagrammatic versions of the axioms of a monoid (associativity and unit axioms).
Following this idea you could easily generalize the construction for structures like magmas and semigroups: you could define a magma object in a monoidal category to be a pair formed by an object $c$ and morphism $c \otimes c \to c$, a semigroup object to be a magma object satisfying the diagrammatic version of the associativity. Applying this magma/semigroup object-construction to the monoidal category of endofunctor I believe you could get the construction you where looking for.
I honestly do not know if such concepts would be useful. Great of the interest in monads arise from the constructions like categories of algebras and the Kleisli category associated to a monad (for what I get this category is what really interest in computer science, because lots of computations can be modelled as morphisms in Kleisli categories for opportune monads defined over $\mathbf{Set}$). These constructions require the whole structure of the monad, so I doubt that you could get something so useful working with semigroupad or magmad (following your notation).
Hope this helps.
p.s. As an aside note I point out the fact that you cannot define a group object in any monoidal category: this is due to problem with the inverse-axiom which requires that your monoidal category has mappings of the form $c \to c \otimes c$, usually called duplicators. This in particular implies that you should not be able to define a groupad in any trivial sense (at least none that I can think of).
The endomorphism monoid of the unit of a monoidal category is commutative. In particular, a monoid has a monoidal structure as one-object category only if it is commutative. Moreover, in that case monoidal structures are given by $f\otimes g=f\circ g$, a strict associator, and an invertible element, considered as an automorphism $m\colon I\otimes I=I\to I$, corresponding to both unitors $\lambda_I$ and $\rho_I$ (the monoidal axioms follow immediately).
For the proof, coherence requires that $\lambda_I=\rho_I\colon I\otimes I\to I$ for the unit $I$ of a monoidal category (this used to be one of the original coherence axioms until shown redundant).
Combined with the fact that $\lambda_X\colon I\otimes X\to X$ and $\rho_X\colon X\otimes I\to X$ are natural, we have an isomorphism $\lambda_I=\rho_I=m\colon I\otimes I\to I$ such that $m\circ I\otimes f=f\circ m=m\circ f\otimes I$ for each endomorphism $f\colon I\to I$. Since $f=(f\circ m)\circ m^{-1}$, we get that $f*g=m\circ f\otimes g\circ m^{-1}$ is a binary operation on endomorphisms of $I$ with unit $\mathrm{id}_I$ that is homomorphic with respect to composition. Eckmann--Hilton therefore applies to conclude $*$ and $\circ$ coincide and are commutative. In particular, if $I\otimes I=I$, then $f\otimes g=m^{-1}\circ(f\circ g)\circ m=f\circ g$ because $m$ commutes with $(f\circ g)$.
Finally, coherence also requires that $a_{I,I,I}\colon I\otimes I\otimes I\to I\otimes I\otimes I$ be the identity since the latter factors as $m^{-1}\otimes I\circ m^{-1}\circ m\circ m\otimes I\colon I\otimes I\otimes I\to I\otimes I\to I\to I\otimes I\to I\otimes I\otimes I$.
Best Answer
The quote you mention does not say that monads are monoids in the usual sense; it says that they are monoids in the monoidal category of endofunctors of $X$. A monoid in a monoidal category (or monoid object, as suggested by Derek Elkins in a comment) is not a monoid; that's a red herring. In general, a monoid object is an oject of a category which is not necessarily concrete, so it need not be any kind of set; in particular, it does not need to have any kind of elements. But seeing a (usual) monoid as a one-object category means precisely that you identify its elements with the arrows of a category; so you can't do that for monoid objects.
You can however, identify such monoids with one-object enriched categories, exactly in the same way that classical monoids can be identified with classical one-object categories (in fact the classical case is just enrichment over sets with the cartesian product).