I want to show $\log_p(\alpha+a_0 \pi)=-\frac{\beta^2}{2} \pi^2+\pi^3 \mathbb{Z}_p[\zeta_p]$, for $p \neq 2$. Consider the cyclotomic extension $\mathbb{Q}_p(\zeta_p)$, its ring of integers $\mathbb{Z}_p[\zeta_p]$ and uniformizer $\pi$.
Also assume that $\log_p(\alpha) \in p \mathbb{Z}_p[\zeta_p]$ for $\alpha \neq 0$. Here $\log_p$ is $p$-adic logarithm.

I want to show $\log_p(\alpha+a_0 \pi)=-\frac{\beta^2}{2} \pi^2+\pi^3 \mathbb{Z}_p[\zeta_p]$, for  $p \neq 2$, $a_0 \in \mathbb{Z}_p[\zeta_p]$.

proceeding,
\begin{align}
\log_p(\alpha+a_0 \pi) &=\log_p(\alpha)+\log_p(1+a_1 \pi), \ \text{where} \ a_1=\frac{a_0}{\alpha} \in \mathbb{Z}_p[\zeta_p].
\end{align}
If we assume $a_1 \in \mathbb{Z}_p[\zeta_p]-\pi \mathbb{Z}_p[\zeta_p]$, then $a_1=\beta+a_2 \pi \mathbb{Z}_p[\zeta_p]$ with $\beta \in \{0,1,2, \cdots, p-1\}$ and $a_2 \in \mathbb{Z}_p[\zeta_p]$.
Then, \begin{align}
\log_p(\alpha+a_0 \pi) &=\log_p(\alpha)+\log_p(1+(\beta+a_2 \pi \mathbb{Z}_p[\zeta] \pi)) \\
&=\left[(\beta+a_2 \pi \mathbb{Z}_p[\zeta] \pi)-\frac{(\beta+a_2 \pi \mathbb{Z}_p[\zeta] \pi)^2}{2}+\frac{(\beta+a_2 \pi \mathbb{Z}_p[\zeta] \pi)^3}{3}-\cdots \right]+\log_p(\alpha)
\end{align}
From the second term of $RHS$, we get $-\frac{\beta^2}{2} \pi^2$,
but how to manipulate and abolish other terms in order to get the conclusion.
If $\log_p(\alpha+a_0 \pi)=-\frac{\beta^2}{2} \pi^2+\pi^3 \mathbb{Z}_p[\zeta_p]$, for  $p \neq 2$ doesn't exactly hold , what would be the nearby relation ?
I mean, I need  $\log_p(\alpha+a_0 \pi)=-\frac{\beta^2}{2}\pi^2+(\cdots \? \cdots)$, for  $p \neq 2$.
Any help please.
 A: I think this boils down to
$$\log_p(1+\beta\pi)\equiv-\frac{\beta^2}2\pi^2\pmod{\pi^3}.$$
Well,
$$\log_p(1+\beta\pi)=\beta\pi-\frac{\beta^2}{2}\pi^2+\cdots+\frac{\beta^p}{p}\pi^p
-\cdots.\tag{*}$$
You want the $\beta\pi$ and $\beta^p\pi^p/p$ terms to "cancel" appropriately.
As you say, $\beta=b+\gamma\pi$ where $b\in\Bbb Z$. Also you can choose the uniformiser
in the extension field to satisfy  $\pi^{p-1}=-p$. In this case
$$\frac{\beta^p}p\pi^p=-(b+\gamma\pi)^p\pi\equiv-b^p\pi
\equiv-b\pi\pmod{\pi^{p+1}}$$
so that $\beta\pi+\beta^p\pi^p/p$ is certainly zero modulo $\pi^3$. I reckon
that the terms in (*) not listed there have $\pi$-valuation at least $3$.
A: Counterexample: $\alpha:=1, a_0:= \dfrac{\zeta_p-1}{\pi}$, so that $\alpha + a_0 \pi =\zeta_p$ and hence $$(*) \qquad \log_p(\alpha+a_0\pi)= \log_p(\zeta_p)=0.$$
But following your notation, we have $a_1=a_0$ and obviously $a_0 \in \mathbb Z_p[\zeta_p]^*$ meaning $\beta \not \equiv 0$ mod $(\pi)$, so if we had
$$ \log_p(\alpha+a_0\pi) \stackrel{?}\equiv -\dfrac{\beta^2}{2}\pi^2 \text{ mod } \pi^3$$
this would contradict $(*)$.
The error in the other answer is explained in my comment to it.
