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. 

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.

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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 $x*y = r$ 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.