Exponential order of unipotent elements in an endomorphism ring of abelian groups

Question

$\DeclareMathOperator\End{End}\newcommand{\Id}{\mathrm{Id}}$Let $E=\End(I)$ be the endomorphism ring of the abelian group $I$.

We have the following statement for $B\in E$, $p$ a prime number and $r$ a positive integer.

Statement $\mathcal{S}(p^r)$: $(\Id+B)^{p^r}=\Id\quad \Leftrightarrow \quad p^r B=0$.

One can prove that the statement $\mathcal{S}(p^r)$ is true if $I$ is a finite cyclic $p$-group or if $I=(\Bbb{Z}/p^{t}\Bbb{Z})^{\oplus k}$ for some $t>r$.

QUESTIONS:

1. Is this true in general?
2. If it is not true in general, for which abelian groups is it true?

Update:

In view of Achim Krauses answer we have to require that $p>2$. Then it is true for the given examples (for the same $p$).

Second update:

I´m leaving the original question, because otherwise the answer of Achim Krause would loose its meaning.

The modified question reads: Assume $I$ is a finite $p$-group and let $E$ be its endomorpism ring. Assume furthermore that $p\mid B^k$ for some positive integer $k$.

Under which additional conditions the following equivalence is true: $(\Id+B)^{p^r}=\Id\quad \Leftrightarrow \quad p^r B=0$

If $I=(\Bbb{Z}/p^{t}\Bbb{Z})^k$ then the statement holds for $p\ge 2k$. Note that in this case $p\mid B^n$ for some $n\ge 1$ implies that $p\mid B^k$.

Answer 1

The statement doesn't even hold in the claimed case: For $p=2,r=1$, $I=\mathbb{Z}/8$, $B=-2\cdot \operatorname{id}$, we have $(\operatorname{id}+B)^2=\operatorname{id}$, but $2B\neq 0$.

EDIT: The question has now edited to claim that the original claim holds for $p>2$. That's not true either: Consider $I = \bigoplus \mathbb{Z} /p^2$, with $p$ summands, and let $(\operatorname{id} +B)$ be the map which cyclically permutes summands. Then $(\operatorname{id} +B)^p =\operatorname{id} $, but $pB\neq 0$.