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.
