Perhaps the comment refers to the fact that in order to generalize rings to structures with noncommutative addition, we cannot simply delete the axiom that addition is commutative, since, in fact, other axioms force addition to be commutative (Hankel, 1867 [1]). The proof is simple: apply both the left and right
distributive law in different order to the term $\rm\:(1\!+\!1)(x\!+\!y),\:$ viz.
$$\rm (1\!+\!1)(x\!+\!y) = \bigg\lbrace \begin{eqnarray}\rm (1\!+\!1)x\!+\!(1\!+\!1)y\, =\, x + \color{#C00}{x\!+\!y} + y\\
\rm 1(x\!+\!y)\!+1(x\!+\!y)\, =\, x + \color{#0A0}{y\!+\!x} + y\end{eqnarray}\bigg\rbrace\Rightarrow\, \color{#C00}{x\!+\!y}\,=\,\color{#0A0}{y\!+\!x}\ \ by\ \ cancel\ \ x,y$$
Thus commutativity of addition, $\rm\:x+y = y+x,\:$ is implied by these axioms:
$(1)\ \ *\,$ distributes over $\rm\,+\!:\ \ x(y+z)\, =\, xy+xz,\ \ (y+z)x\, =\, yx+zx$
$(2)\ \, +\,$ is cancellative: $\rm\ \ x+y\, =\, x+z\:\Rightarrow\: y=z,\ \ y+x\, =\, z+x\:\Rightarrow\: y=z$
$(3)\ \, +\,$ is associative: $\rm\ \ (x+y)+z\, =\, x+(y+z)$
$(4)\ \ *\,$ has a neutral element $\rm\,1\!:\ \ 1x = x$
In order to state this result concisely, recall that a SemiRing is
that generalization of a Ring whose additive structure is relaxed
from a commutative Group to merely a SemiGroup, i.e. here the only
hypothesis on addition is that it be associative (so in SemiRings,
unlike Rings, addition need not be commutative, nor need every
element $\rm\,x\,$ have an additive inverse $\rm\,-x).\,$ Now the above result may
be stated as follows: a semiring with $\,1\,$ and cancellative addition
has commutative addition. Such semirings are simply subsemirings
of rings (as is $\rm\:\Bbb N \subset \Bbb Z)\,$ because any commutative cancellative
semigroup embeds canonically into a commutative group, its group
of differences (in precisely the same way $\rm\,\Bbb Z\,$ is constructed from $\rm\,\Bbb N,\,$
i.e. the additive version of the fraction field construction).
Examples of SemiRings include: $\rm\,\Bbb N;\,$ initial segments of cardinals;
distributive lattices (e.g. subsets of a powerset with operations $\cup$ and $\cap$;
$\rm\,\Bbb R\,$ with + being min or max, and $*$ being addition; semigroup semirings
(e.g. formal power series); formal languages with union, concat; etc.
For a nice survey of SemiRings and SemiFields see [2]. See also Near-Rings.
[1] Gerhard Betsch. On the beginnings and development of near-ring theory.
pp. 1-11 in:
Near-rings and near-fields. Proceedings of the conference
held in Fredericton, New Brunswick, July 18-24, 1993. Edited by Yuen Fong,
Howard E. Bell, Wen-Fong Ke, Gordon Mason and Gunter Pilz.
Mathematics and its Applications, 336. Kluwer Academic Publishers Group,
Dordrecht, 1995. x+278 pp. ISBN: 0-7923-3635-6 Zbl review
[2] Hebisch, Udo; Weinert, Hanns Joachim. Semirings and semifields. $\ $ pp. 425-462 in: Handbook of algebra. Vol. 1. Edited by M. Hazewinkel.
North-Holland Publishing Co., Amsterdam, 1996. xx+915 pp. ISBN: 0-444-82212-7
Zbl review,
AMS review
Let $f = \sum r_i x^i$ and $g = \sum s_j x^j$ be two polynomials. I claim that in the product $fg$ there is a term whose coefficient has the form $r_i s_j$ for some $i, j$. This suffices to prove the claim under the assumption that $R$ has no zero divisors, and reduces the claim to the following straightforward geometric argument:
Since $f, g$ both have finitely many terms, we may assume WLOG that $X$ is finite, say $|X| = n$. Let
$$\text{supp}(f) = \{ i \in \mathbb{Z}^n : r_i \neq 0 \}$$
denote the support. Then $\text{supp}(f)$ and $\text{supp}(g)$ are two finite sets of points in $\mathbb{Z}^n \subset \mathbb{R}^n$. Let $H$ be a hyperplane in $\mathbb{R}^n$ such that none of its translates passes through two or more points of either $\text{supp}(f)$ or $\text{supp}(g)$ (a generic hyperplane will have this property). Let $v \in \mathbb{R}^n$ be a vector orthogonal to $H$, and say that a point in a subset of $\mathbb{R}^n$ is extremal if $\langle v, - \rangle$ attains a maximum there. By construction, $\text{supp}(f)$ and $\text{supp}(g)$ have unique extremal points $i_0, j_0$ (if there is more than one extremal point then some translate of $H$ passes through all of them).
Now, $\text{supp}(fg)$ is contained in the Minkowski sum $\{ i + j : i \in \text{supp}(f), j \in \text{supp}(g) \}$. Furthermore, $\langle v, i + j \rangle = \langle v, i \rangle + \langle v, j \rangle$, from which it follows that
$$\langle v, i + j \rangle \le \langle v, i_0 + j_0 \rangle$$
with equality iff $i = i_0, j = j_0$. In particular, $i + j \neq i_0 + j_0$ unless $i = i_0, j = j_0$. Hence the coefficient of $x^{i_0 + j_0}$ in $fg$ is $r_{i_0} s_{j_0}$ and the conclusion follows.
The geometric picture is visualizable when $n = 2$. Here imagine two collections of points in the plane, and take e.g. "leftmost points" (although you may have to tilt the plane slightly if there is more than one leftmost point in each collection).
Best Answer
The $f$ that you defined, a priori, can only be a set-theoretic bijection $R\to S$, because even if you named two operations on $S$, we don't know the structure of $S$ (you didn't give axioms to define how the operations work on $S$).
This situation could be what you were thinking about: suppose that you have a ring $(R,\cdot,+,1_R,0_R)$ and a set $S$. If $f:S\to R$ is a bijection (viewing $R$ as a set), we can actually give $S$ the structure of a ring. We can define the ring $(S,\otimes,\oplus,1_S,0_S)$ like follows ($s,s'\in S$):
$s\otimes s'$ is the (unique because $f$ is bijective) element $t\in S$ such that $f(t)=f(s)\cdot f(s')$; similarly, $s\oplus s'$ is the element $t\in S$ such that $f(t)=f(s)+ f(s')$; and obviously $0_S:=f(0_R)$ and $1_S:=f(1_R)$.
You may verify that, with this definition for the operations, $S$ is a ring. For example, let us verify the distributive property; we will denote with $t_r$, for any $r\in R$, the element of $S$ for which $f(t_r)=r$. $$s_1\otimes(s_2\oplus s_3)=t_{f(s_1)\cdot f(s_2\oplus s_3)}=t_{f(s_1)\cdot f(t_{f(s_2)+f(s_3)})}=t_{f(s_1)\cdot (f(s_2)+f(s_3))}=t_{(f(s_1)\cdot f(s_2))+(f(s_1)\cdot f(s_3))}$$ $$=t_{f(s_1)\cdot f(s_2)}\oplus t_{f(s_1)\cdot f(s_3)}=(s_1\otimes s_2)\oplus(s_1\otimes s_3).$$ (Justifying all the equalities may seem confusing at first, but they are all immediate from the definitons and the distributivity of $R$).
PS: notice that $f$ must be defined $S\to R$, and not viceversa; intuitively, you can convince yourself that it is the right direction thinking that $S$ can be given a ring structure if it can be "inserted" in a ring (in fact the hypothesis of $f$ surjective could be replaced with a weaker one, while the injectivity is fundamental).