Is there a valid multiplication for any choice of identity in $C(\mathbb{R})$? Let $C(\mathbb{R})$ be the ring of continuous functions from $\mathbb{R}$ to $\mathbb{R}$. Its identity with the usual multiplication is $1(x) = 1$.
I have two related questions. Firstly, when we choose our multiplication to be function composition instead of the usual function multiplication, we "almost" get a ring, in that every axiom except distributivity from the left holds. This multiplication would have identity $1(x) = x$. Composition clearly won't work, but is there a valid multiplication on $C(\mathbb{R})$ which has $1(x) = x$ as its identity?
Secondly, if I specify a function $f(x)$  when will there be a valid multiplication on $C(\mathbb{R})$ with $f$ as its identity?
Thanks for your time!
 A: There is an answer to this question that works, however, it is not very satisfying.
If I understand your question correctly, you consider $C(\mathbb{R})$ together with pointwise addition every time. So it is an Abelian group. Suppose we have chosen $f$ to be the identity for multiplication. For $f=0$, i.e. the everywhere zero function, the task is obviously impossible (the only unital ring with $1=0$ is the zero ring, i.e. $R=\{0\}$). So assume that $f \neq 0$ (i.e. at least one of its values is nonzero).
Let us consider an additional structure - multiplication by real scalars. Thus we have $(C(\mathbb{R}),+,-,0, (-\cdot r)_{r\in \mathbb{R}})$, a real vector space. Denote its dimension $\kappa$ (dimension is wrt "algebraic", i.e. Hamel, basis). Choose a basis $B$ such that $f \in B$ (this is possible to do by Zorn's lemma; note that here we are using the fact that $f \neq 0$).
Now let us fix an $\mathbb{R}$-algebra of dimension $\kappa$, for example, $A=\mathbb{R}[\{X_\alpha \; | \; \alpha \in \kappa\}]$ - i.e. real polynomials over $\kappa$ many indeterminates $X_\alpha, \alpha \in \kappa$ (every such polynomial, of course, uses only finitely many of them). It is easy to see that 
$$ C=\{X_\alpha^n \; | \; \alpha \in \kappa, n \in \mathbb{N}\} \cup \{1\} $$
is a basis of the underlying vector space of the algebra.
Now $B\setminus\{f\}$ and $C \setminus \{1\}$ are of the same cardinality ($=\kappa$), hence we can consider a bijection $\varphi:B \rightarrow C$ such that $\varphi(f)=1$. Then $\varphi$ can be uniquely extended to an isomorphism of vector spaces $\psi: C(\mathbb{R})\rightarrow \mathbb{R}[\{X_\alpha \; | \; \alpha \in \kappa\}]$ (it is a bijection between bases). So we have an isomorphism with $\psi(f)=1$, the unit of $A$.
Now define operation $\odot$ on $ C(\mathbb{R})$ by 
$$g\odot h \stackrel{def}= \psi^{-1}(\psi(g)\cdot \psi(h)),$$
where $\cdot$ denotes the multiplication in the algebra $A$. Then $(C(\mathbb{R}),+,-,
0,\odot,f)$ is a commutative ring with unit $f$, isomorphic to $A$. (basically, what we did is that we "stole" the ring structure of $A$ by translating it through $\psi$ to $C(\mathbb{R})$). 
Of course, this structure does not have much in common with the "natural structure" on $C(\mathbb{R})$, apart from respecting the real vector space structure.
