There is a straightforward approach to proving that isomorphism of monoids $\ \text{End}(X\ T)\ $ and $\ \text{End}(\Bbb R\,\ T_E)\ $ implies homeomorphism of $\ (X\ T)\ $ and Euclidean space $\ (\Bbb R\,\ T_E):\ $
let $\ (M\ \circ\ J)\ $ be an arbitrary abstract monoid. One may associate with this monoid a set of "points" $\ C\ $ of "constant" elements, as @YCor did in his first comment, as the left-absorbing elements $\ c\in C,\ $ where
$$ \forall_{f\in M}\quad c\circ f=c $$
Then one selects purely algebraic (monoidal) notions which have the respective topological meaning in $\ (\Bbb R\,\ T_E).\ $ You need just a couple of such notions.
Then you force on the abstract monoid $\ (M\ \circ\ J)\ $ the purely algebraic (monoidal) axioms that induce the properties of reals. That's all.
For instance one may apply idempotents $\ i\in\mathcal I\subseteq M,\ $ where $\ i\circ i=i. \ $ Then the critical notion to me is what I've defined and call universally closed idempotents or uc-morphism $\ i\in\mathcal I,\ $ which satisfy:
$$ \forall_{f\in M}\,\exists_{p\in M}\quad i\circ f\circ i\circ p\,\ =\,\ i\circ p $$
Observe that we have a canonical map
$$ \pi: \mathcal I\to 2^C, $$
where
$$ \forall_{i\in\mathcal I}\quad \pi(i)\ := \ \{c\in C:\ i\circ c=c\} $$
This has nice properties... etc.
(After I saw the OP question, I hesitated... and I can remove my post, no sweat.)