This is a long comment rather than a complete answer. But before writing it let me insert that I don't agree that universal algebra is the study of varieties. It is the study of algebraic structures. (In my universe, universal algebra is synonymous with algebra.)
Nevertheless, it is clear that this question is about varieties, so I will write about them.
Let $Var$ be the category of varieties. The morphisms between objects are the homomorphisms between the clones of the varieties. One object in $Var$ is the variety ${\mathcal V}_{\mathbb R}$ generated by the algebra whose underlying set is $\mathbb R$ and whose operations are all continuous operations on $\mathbb R$.
Now let $\mathcal U$ be any other variety. Each morphism $\mathcal U\to {\mathcal V}_{\mathbb R}$ corresponds to a way of equipping $\mathbb R$ with compatible continuous operations defining a $\mathcal U$-structure on $\mathbb R$.
If you want to know `what kinds of algebras are compatible with the reals', then you want to know the principal ideal in $Var$ defined by ${\mathcal V}_{\mathbb R}$: i.e. $({\mathcal V}_{\mathbb R}]:=\{{\mathcal U}\;|\;\exists \varphi(\varphi\colon {\mathcal U}\to {\mathcal V}_{\mathbb R}\;\textrm{is a hom})\}$. Moreover, if you want to know all ${\mathcal U}$-structures on $\mathbb R$, you want to know all homomorphisms $\varphi\colon {\mathcal U}\to {\mathcal V}_{\mathbb R}$.
(The last sentence of the previous paragraph shows that a better object than $({\mathcal V}_{\mathbb R}]$ for this question is the slice category $Var/{\mathcal V}_{\mathbb R}$, since this identifies not just the types of algebras definable on $\mathbb R$ by continuous operations, but also their realizations.)
Taylor showed that there is no algorithm to determine if a finitely presented variety belongs to $({\mathcal V}_{\mathbb R}]$. He does this by interpreting the undecidable satisfiability problem for Diophantine equations into this problem.
On the other hand, Taylor shows that there is an easy algorithm to determine whether a variety $\mathcal U$ defined by a finite set $\Sigma$ of simple equations belongs to $({\mathcal V}_{\mathbb R}]$. (An equation is simple if each side has at most one function symbol.) He shows that such ${\mathcal U}$ belongs to $({\mathcal V}_{\mathbb R}]$ iff $\mathcal U$ has a 2-element model.
From Taylor's work it is easy to see that it is possible to equip $\mathbb R$ with a continuous commutative binary operation, since $x*y=y*x$ is a simple equation that has a $2$-element model. But Taylor's work does not tell us all ways of equipping $\mathbb R$ with a continuous commutative binary operation.
Taylor's work says nothing about the associative law, because it is not simple.
Regarding Q2.1: There are continuumly many continuous, associative, binary operations on $\mathbb R$. There can't be more, and it is not hard to construct this many. For a trivial construction of many continuous associative operations, define $x*_ry = r$ for different $r\in \mathbb R$. Less trivially, select an interval $[a,b]$ and define $f(x) = \min(\max(x,a),b)$. The operation $x*y = \max(f(x),f(y))$ is continuous, associative, depends on both variables, and has range $[a,b]$. You get a lot of these by varying $a$ and $b$. From some given continuous, associative, binary operations you can often generate more by conjugating by continuous permutations of $\mathbb R$. ($x*_{\pi}y:=\pi^{-1}(\pi(x)*\pi(y))$.)
Regarding Q2.2: There are obstacles to constructing a third associative binary operation from two given ones, if you intend to construct the third by some form of composition. For example, the projections $\ell(x,y)=x$ and $r(x,y)=y$ are both associative, but the only binary operations you can get from these by composition are $\ell(x,y)$ and $r(x,y)$ themselves. Similarly, $\max(x,y)$ and $\min(x,y)$ are associative, but the only binary operations in the clone they generate are $\max, \min, \ell, r$. So any general construction would have to trivialize in these situations.
Nevertheless, here is something. Suppose you are given continuous, associative, binary operations $x+y$ and $x*y$. Assume in addition that (i) addition is commutative, and (ii) multiplication distributes over addition on both sides. Then $x\circ y:= x+y+x*y$ is a third continuous, associative, binary operation.