Assume $f \in \Bbb Z[x]$ is a monic polynomial, s.t for every commutative ring $R$, the solutions of $f(x)=0$ in $R$ can be endowed with an abelian group structure that is functorial respect to $R$.
Some examples of f include $ f(x)=x^n-1, f(x)=x$ , with the group structure inherited from the additive group or the unit group. But there is an example with new group structure: $ f(x)=x^2-x$ with group operation given by $(x,y) \mapsto (x-y)^2$.
So are there other types of such polynomial？How to classify all such $f$？
A neccesary condition is $f$ has root in every $R$ hence $f $ must have a linear factor.
This is really only a place to start and not an answer, but it's too long to be a comment.
As I said in my first comment, you want to put a Hopf algebra structure on $A = \Bbb Z[x]/(f)$ which is both commutative and cocommutative. That is, you need ring homomorphisms: \begin{align*} e : \Bbb Z&\to A&\text{unit}\\ \epsilon : A&\to\Bbb Z&\text{counit}\\ \mu : A\otimes_{\Bbb Z}A&\to A&\text{multiplication}\\ \Delta : A&\to A\otimes_{\Bbb Z}A&\text{comultiplication}\\ S : A&\to A&\text{antipode} \end{align*} such that the following conditions hold: \begin{align*} \mu\circ(id_A\otimes\mu) &= \mu\circ(\mu\otimes id_A)&\text{associativity}\\ id_A &= \mu\circ(id_A\otimes e)\circ(A\xrightarrow{\sim} A\otimes\Bbb Z)&\text{identity}\\ \mu &= \mu\circ\tau&\text{commutativity}\\ (id_A\otimes\Delta)\circ\Delta &= (\Delta\otimes id_A)\circ\Delta&\text{coassociativity}\\ id_A &= (A\otimes\Bbb Z\xrightarrow{\sim}A)\circ(id_A\otimes\epsilon)\circ\Delta&\text{coidentity}\\ \tau\circ\Delta &= \Delta&\text{cocommutativity}\\ \Delta\circ\mu &= \left(\mu\otimes\mu\right)\circ\left(id_A\otimes\tau\otimes id_A\right)\circ\left(\Delta\otimes\Delta\right)&\text{compatibility}\\ \Delta\circ e &= (e\otimes e)\circ(\Bbb Z\xrightarrow{\sim}\Bbb Z\otimes\Bbb Z)\\ \epsilon\circ\mu &= (\Bbb Z\otimes\Bbb Z\xrightarrow{\sim}\Bbb Z)\circ(\epsilon\otimes\epsilon)\\ e\circ\epsilon &=\mu\circ(S\otimes id_A)\circ\Delta&\text{antipode},\\ \end{align*} where $\tau : A\otimes A\to A\otimes A$ denotes the "switching coordinates" map $x\otimes y\mapsto y\otimes x$. $A$ is already a commutative $\Bbb Z$-algebra, so the first three conditions do not add anything new. The condition you mentioned, that $f$ has a linear factor, arises immediately as a result of existence of a counit, since a map $A\to\Bbb Z$ must send [the class of] $x$ in $A$ to an integer root of $f$.
If you write out the potential comultiplication on $x$ as $\Delta(x) = \sum_{i,j\geq 0} n_{i,j} x^i\otimes x^j$ and use the other conditions, you might get some things. For example, $$ (id_A\otimes\Delta)\circ\Delta(x)= (\Delta\otimes id_A)\circ\Delta(x) $$ implies that $$ \sum_{i,j\geq 0}\left[n_{i,j} x^i\otimes\left(\sum_{k,\ell\geq 0} n_{k,\ell} x^k\otimes x^\ell\right)^j\right]= \sum_{i,j\geq 0}\left[n_{i,j} \left(\sum_{k,\ell\geq 0} n_{k,\ell} x^k\otimes x^\ell\right)^i\otimes x^j\right], $$ and $$ id_A(x) = (A\otimes\Bbb Z\xrightarrow{\sim}A)\circ(id_A\otimes\epsilon)\circ\Delta(x) $$ implies that $$ x = \sum_{i,j\geq 0} n_{i,j}\epsilon(x)^j x^i, $$ so that $\sum_{j\geq 0} n_{1,j}\epsilon(x)^j = 1,$ and for $i\geq0$, $i\neq 1$, you have $\sum_{j\geq 0} n_{i,j}\epsilon(x)^j = 0.$ $\tau\circ\Delta = \Delta$ implies that $n_{i,j} = n_{j,i}$, and so on.
Perhaps a good place to start would be by fixing $\epsilon$ (or fixing the distinguished root of $f$, if you like): say that $\epsilon(x) = 0$. Then $f$ has no constant term, and some of the above results become particularly simple. You get that $$ 1 = \sum_{j\geq 0} n_{1,j}\epsilon(x)^j = n_{1,0}, $$ and for $i\geq0$, $i\neq 1$, you have $$ 0 = \sum_{j\geq 0} n_{i,j}\epsilon(x)^j = n_{i,0}. $$
Of course, it remains to be seen how you can interpret the existence of such a $\Delta$ (and $S$) as a condition on $f$ itself.