With your additional hypothesis that $x$ does not lie in the prime ideal $(1- \zeta)$, the question becomes much easier, because of its now "local" nature. To simplify notations, put $\pi=1 - \zeta $ throughout and consider the ring $A = \mathbf Z_p[\zeta]$, where $\mathbf Z_p$ denotes the $p$-adic integers. The group $U$ of units (= invertible elements) of $A$ is endowed with a descending filtration $U > U_1 > ... > U_n > ...$, where $U_n = 1+ (\pi^{n})$ . The residual field $k = A/(\pi)$ is isomorphic to $\mathbf F_p$ because $p$ is totally ramified, and the residue map $A \to \mathbf F_p$ induced by $\zeta \to 1$ gives rise to an isomorphism $U/U_1 \cong \mathbf F_p^{*}$ (multiplicative group), whereas for $n\ge 1$ the map $u \to u - 1$ gives rise to $U_n/U_n+1 \cong (\pi^{n})/(\pi^{n+1})\cong \mathbf F_p$ (additive group) (see e.g. Cassels-Fröhlich, chap. 1, propos. 4). One gets in particular an exact sequence $0 \to \mathbf F_p \to U/U_2 \to \mathbf F_p^{*} \to 1$, which is split because the two extreme terms have coprime orders. Given the explicit isomorphisms above, this shows that $U/U_2 = <\zeta>.\mathbf Z_p$ , hence for any $u\in U$ , a certain $u.\zeta^{r}$ will be congruent to $a$ mod $(\pi^{2})$, with $a \in\mathbf Z_p$. Coming back to our $x \in \mathbf Z[\zeta], \notin (\pi)$, and using the density of $\mathbf Z$ in $\mathbf Z_p$, a certain $x.\zeta^{r}$ will be will be congruent to $b$ mod $(\pi^{2})$, with $b \in\mathbf Z$ as desired.
The $p$-adics were unknown at Kummer's time, but the above proof can be adapted to be (probably)more in Kummer's spirit. Consider the ring $B = \mathbf Z[\zeta]= \mathbf Z[\pi] $. Every element $x \in B$ can be uniquely written under the form $x = a_0 + a_1\pi + ... + a_{p - 1}\pi^{p-1}$, with $a_n \in \mathbf Z$. Analogously to above, the natural exact sequence $ 0 \to (\pi)/(\pi^{2}) \to B/(\pi^{2}) \to B/(\pi) \cong \mathbf F_p \to 0$ gives rise to another exact sequence $ 0 \to (\pi)/(\pi^{2})\cong \mathbf F_p \to (B/(\pi^{2}))^{*} \to (B/(\pi))^{*} \cong \mathbf F_p^{*} \to 0$, where $(.)^*$ denotes the units. The hypothesis that $x \notin (\pi)$ means that $a_0$ is not a multiple of $p$, hence by the definition of the residue map, that $x$ mod $(\pi)$ $\in (B/\pi)^{*}$. Analogously to above, one can conclude that a certain $x.\zeta^{r}$ will be will be congruent to $b$ mod $(\pi^{2})$, with $b \in\mathbf Z$ as desired.
NB: This second proof is just a formalization of the one given by @Starfall.