Which polynomials make $\mathbb{R}$ into a monoid? 
Question. Can we describe the set $$\{P : \mathbb{R}^2 \rightarrow \mathbb{R}, e:\mathbb{R}\mid P\; \mbox{is a bivariate polynomial}, P \mbox{ is associative}, P(e,x) = P(x,e) = x\}$$
  explicitly?

Noting that $+$ is induced by the polynomial $x+y$, and $\times$ is induced by the polynomial $xy$, hence examples include $(+,0)$ and $(\times,1)$. I'll bet there's heaps of others, though.
 A: Let $K$ be any field and $P(x,y)\in K[x,y]$ be a polynomial and $e\in K$ be such that $P(x,e)=x$, $P(e,y)=y$, and $P(P(x,y),z)=P(x,P(y,z))$.  (Note: We require that these be equalities of polynomials, not just of functions, which makes a difference if $K$ is finite.)  Then $P$ is conjugate to either addition or multiplication by an affine map $K\to K$.
Indeed, suppose you have any such polynomial $P$ with identity $e$.  After conjugating by a translation $K\to K$, we may assume $e=0$.  The conditions $P(x,0)=x$ and $P(0,y)=y$ now say that $P(x,y)$ has the form $$P(x,y)=x+y+\text{terms divisible by $xy$}.$$
Let $\overline{K}$ be an algebraic closure of $K$.  Note that for all but finitely many $a\in\overline{K}$, the polynomials $P(x,a)\in \overline{K}[x]$ and $P(a,y)\in\overline{K}[y]$ are nonconstant.  In particular, they have roots in $\overline{K}$.  This means that all but finitely many elements of $\overline{K}$ have both a left inverse and a right inverse (which then must be equal) in the monoid structure on $\overline{K}$ given by $P$.
Note moreover that for any such invertible $a\in\overline{K}$, $x\mapsto P(x,a)$ is a polynomial in $x$ which has a polynomial inverse (namely $x\mapsto P(x,a^{-1})$), and hence must have degree $1$.  If $n$ is maximal such that $P$ has a term divisible by $x^n$, then for all but finitely many $a\in\overline{K}$, $P(x,a)$ has a nonzero coefficient of $x^n$.  In particular, we can choose such an $a$ which is invertible in our monoid.  It follows that we must have $n=1$.  Similarly, the largest $n$ such that $P$ has a term divisible by $y^n$ must also be $1$.
We conclude that $P$ has the form $x+y+cxy$ for some $c\in K$.  If $c=0$, we have addition.  If $c\neq 0$, then conjugating by $x\mapsto x/c$ turns $P$ into $x+y+xy$, which is conjugate to multiplication by $x\mapsto x-1$.
