Any element of $\mathbf{Z}[\xi]$ is congruent to an integer modulo $(1-\xi)^2$ if multiplied by a suitable power of $\xi$
This is false - $ 1 - \xi $ can never be multiplied by a power of $ \xi $ to be an integer modulo $ (1 - \xi)^2 $. Let $ \mathfrak p = (1 - \xi) $ throughout the post.
To see this, note that $ p \equiv 0 \pmod{\mathfrak p^2} $ since the ideal $ (p) $ totally ramifies as $ (p) = \mathfrak p^{p-1} $, so if $ \xi^k (1 - \xi) $ is an integer modulo $ \mathfrak p^2 $, without loss of generality it can be taken to be in the set $ \{ 0, 1, \ldots, p-1 \} $. It cannot be $ 0 $, because that would imply $ 1 - \xi \in \mathfrak p^2 $ and $ \mathfrak p = \mathfrak p^2 $, but $ 1 - \xi $ is not a unit: it has norm $ \pm p $. It cannot be anything else in this set, because $ \mathbf Z \cap \mathfrak p = p\mathbf Z $, therefore all of the nonzero integers in the set are invertible modulo $ \mathfrak p $, and thus modulo $ \mathfrak p^2 $. Thus, such a congruence would imply that $ 1 - \xi $ is invertible modulo $ \mathfrak p^2 $, which is absurd, since the ideals $ (1 - \xi) = \mathfrak p $ and $ \mathfrak p^2 $ are not coprime.
Perhaps you dropped some hypothesis in your statement, which should be: "Any unit $u$ in $\mathbf Z[\zeta]$ can be multiplied, etc." By Dirichlet's unit theorem (or by direct computation), $u$ is of the form $\zeta^r .v$, where $v$ is a unit in $\mathbf Z[\zeta +\zeta ^{-1}]$ (= the ring of integers of the maximal totally real subfield). But $(1-\zeta)(1-\zeta ^{-1}) = 2 -(\zeta +\zeta ^{-1})$, and $(1-\zeta)/(1-\zeta ^{-1})$ is a unit, hence taking quotients mod $(1-\zeta)(1-\zeta ^{-1})$ shows that every element of $\mathbf Z[\zeta +\zeta ^{-1}]$ is congruent to a rational number mod $(1-\zeta)^2$ QED
I guess that this property is a preliminary in the proof of the so called Kummer lemma, in which case you could fine more information in Washington's book, §5.6.
Answer to the bounty question:
Denote $ R = \mathbf Z[\xi]/(1 - \xi)^2 $. We make two observations: the map $ \mathbf Z \to R $ has kernel $ p\mathbf Z $, so it descends to an embedding $ \mathbf Z/p \mathbf Z \to R $, and since an integer is invertible modulo $ (1 - \xi)^2 $ if and only if it is not divisible by $ p $ (this was shown in my other answer), it follows that this embedding gives an embedding of groups $ H = (\mathbf Z/p \mathbf Z)^{\times} \to R^{\times} $. On the other hand, $ \xi $ is certainly in $ R^{\times} $, and its order divides $ p $, so it is either congruent to $ 1 $ or has order $ p $ in $ R^{\times} $. The former is impossible since it implies that $ 1 - \xi $ is divisible by $ (1 - \xi)^2 $, so $ \xi $ is an element of order $ p $. Then, the subgroup $ K = \{ x \in R^{\times} : x = \xi^i z, z \in \mathbf Z \} $ is the product of the subgroups $ H $ and $ \langle \xi \rangle $, which intersect trivially since their orders are coprime. It follows that $ |K| = |H| |\langle \xi \rangle| = p(p-1) $.
On the other hand, $ N((1 - \xi)^2) = N(1 - \xi)^2 = p^2 $, so $ |R| = p^2 $, and the submodule $ (1 - \xi) / (1 - \xi)^2 $ of $ R $ has cardinality $ p $, which consists entirely of non-invertible elements in $ R $. Thus, $ |R^{\times}| = p^2 - p = p(p-1) $, and from this it follows that $ K = R^{\times} $. The statement follows upon writing $ x \equiv \xi^i z \pmod{(1 - \xi)^2} $ for $ x \notin (1 - \xi) $ and multiplying both sides by $ \xi^{p-i} $.