As your opening paragraph states, in a ring of (univariate) polynomials over a field $F$, say $F[x]$, each nonzero polynomial $p(x)$ has exactly one monic associate (a multiple $cp(x)$ where $c$ is a unit in $F[x]$, in this case a nonzero element of $F$).
In a similar polynomial ring $R[x]$ over an arbitrary integral domain $R$ this is no longer true, not because a nonzero polynomial $p(x)$ could have more than one monic associate but because it may have no monic associate at all.
The situation with integral domain $\mathbb Z$ illustrates this. Consider a nonzero first degree polynomial, say $p(x) = mx + b$ with $m,b \in \mathbb Z$. What would it mean for a polynomial to be a monic associate of $p(x)$? We would need to find a unit in $\mathbb Z[x]$ such that multiplying it by $p(x)$ gives us the desired monic associate.
But the only units in $\mathbb Z[x]$ are $\pm 1$. So $p(x)$ has a monic associate only if $m=\pm 1$, so the sought polynomial without a monic associate could be as simple as $p(x) = 2x$.
What is true for integral domain $R$ is that when a polynomial $p(x) \in R[x]$ has a monic associate, it has exactly one monic associate (uniqueness). This follows (with perhaps a little thought) from observing that the units of $R[x]$ are precisely the units of $R$ (assuming the usual inclusion in $R[x]$). It then happens that a polynomial $p(x) \in R[x]$ has a monic associate if and only if the leading coefficient of $p(x)$ is a unit of $R$. Of course this happens for all nonzero polynomials in $R[x]$ only when $R$ is a field (so that all nonzero leading coefficients are units). We've now circled back to the beginning of your problem's setup.