Is $\overline{\langle2\rangle\cdot(4\Bbb Z+1)}=\langle2\rangle\cdot(4\Bbb Z_2+1)$ in $\Bbb Z_2$?
$\langle2\rangle$ is the set of powers of $2$ and $\cdot$ is the straightforward dot product.
I get that $\overline{(4\Bbb Z+1)}=(4\Bbb Z_2+1)$ where $\Bbb Z_2$ is the 2-adic integers. But what about if we introduce powers of $2$ as factors? Tentatively I would say it seems the given proposition follows because $2b$ is simply the centre of the ball $b$ but then I wonder about the number $0$ which would be in one but not the other, so I'm confused!
Maybe then:
$\overline{\langle2\rangle\cdot(4\Bbb Z+1)}=\overline{\langle2\rangle}\cdot(4\Bbb Z_2+1)$ in $\Bbb Z_2$?
There's no reason to assume that the elementwise product of sets plays particularly well together with taking closure.
For example, in good old $\mathbb R$ consider the set $A=\mathbb Z\cup\{1/n\mid n\in\mathbb N_+\}$. This is a closed countable set, so $\overline A\cdot \overline A$ is countable. However $A\cdot A$ gives you $\mathbb Q$, so $\overline{A\cdot A}$ is much larger than $\overline A\cdot \overline A$.
So you shouldn't hope to compute your product just by symbolic guesswork -- if it has a nice description it will be more a matter of luck.
In your case you're right that $0$ is in $\overline{\langle 2\rangle\cdot(4\mathbb Z+1)}$, but is not in $\langle 2\rangle\cdot(4\mathbb Z_2+1)$, so those sets are definitely not equal.
It does look to me like $\overline{\langle 2\rangle\cdot(4\mathbb Z+1)}=\overline{\langle 2\rangle}\cdot(4\mathbb Z_2+1)$, but to be sure of that you need an ad-hoc proof.