A binary operation, closed over the reals, that is associative, but not commutative
We can define $x \oplus y=y$. Then $(x \oplus y) \oplus z =z= x \oplus (y \oplus z)$ but $y=x \oplus y \neq y \oplus x=x$
If you already know that matrix multiplication is associative, but not commutative, then you can just choose your favorite bijection $f:\mathbb R\rightarrow M_2(\mathbb R)$, since the two sets are equinumerous. Then, define $a\oplus b = f^{-1}(f(a)f(b))$ to get an operation on $\mathbb R$ which is associative, but not commutative. If you want to have inverses as well, then you can replace $f$ with your favorite map from $\mathbb R$ to the invertible $2\times 2$ matrices.
In general, without more structure, you are equivalently asking, "Is there any associative, but not commutative operation defined on a domain of cardinality $|\mathbb R|$?" since the role of $\mathbb R$ in the question is nothing more than a set.
Batominovski has another answer in the comments to this question. I will type up the checking:
Our candidate is $x\circ y=|x|y$. Then:
- Associative? We have $(x\circ y)\circ z=(|x|y)\circ z=||x|y|z=|xy|z.$ On the other hand, $x\circ(y\circ z)=x\circ(|y|z)=|x||y|z=|xy|z.$
- Non-commutative? $x\circ y=|x|y\not=|y|x=y\circ x$.
So this solution of Batominovski's fits the bill.