Let $a = (a_n)_{n\in\mathbb{Z}}$ where $a_n\geq 2$ are natural numbers. We define the $a$-adic numbers $\Omega_a$ just like one would define $r$-adic numbers for natural number $r\geq 2$, that case corresponding to a constant sequence $(r)_{n\in\mathbb{Z}}$, an element $x\in \Omega_a$ being a sequence of natural numbers $x = (x_n)_{n\in\mathbb{Z}}$ with $0\leq x_n < a_n$ and there is $m$ such that $x_n = 0$ for all $n\leq m$. And in the same way we define addition: $z = x+y$ iff $z_n+t_na_n = x_n+y_n+t_{n-1}$ for some $t_n\geq 0$ and if $m$ is least so that $x_m$ or $y_m$ are non-zero, then $t_n = 0$ for $n< m$. Then $(\Omega_a, +)$ is a group.
Contrary to $r$-adic numbers we can have elements of finite order in $\Omega_a$. For which $a$ is $\Omega_a$ a torsion-free group?
Edit: Solution for the categorically inept like me:
If $\Omega_a$ has torsion then $px = 0$ for some non-zero $x$ and prime $p$ (since if $mx = 0$ we can recursively factorize $m$). We can assume $x_0\neq 0$ and $x_n = 0$ for $n<0$. We have $px_n + t_{n-1} = t_na_n$ for $n \geq 0$ where $t_{-1} = 0$. If it were $p|t_0$, then $px_0 < t_0a_0$, contradiction. So $p|a_0$. But if it were $p|a_k$ for some $k>0$ then $p|t_n$ for all $n < k$, contradiction. So $p$ doesn't divide $a_1, a_2, ...$.
Conversely in the above situation, let $x_0 = a_0/p \neq 0$ and recursively solve $px_n +t_{n-1} = t_na_n$ as follows: solve $t_{n-1} = t_n'a_n \pmod{p}$ so that $px_n' +t_{n-1} = t_n'a_n$. Write $x_n' = x_n+sa_n$ where $0\leq x_n < a_n$ and $t_n = t_n'-sp$. Then $x = (x_n)$ is such that $px = 0$.
Let $\Omega_a^m$ denote the subgroup of $\Omega_a$ consisting of elements with $x_n=0$ for all $n<m$. Then $\Omega_a^m$ is the inverse limit of the quotient maps between the groups $\mathbb{Z}/(b_n)$ where $b_n=a_{m}a_{m+1}\dots a_{m+n}$. These groups split into their torsion with respect to each prime and so $\Omega_a^m$ is the product of the inverse limits with respect to each prime. This means $\Omega_a^m$ is a product of groups $\mathbb{Z}/(p^{n_p})$ where $n_p$ is the total number of factors of $p$ in all the numbers $a_n$ for $n\geq m$, or $\mathbb{Z}_p$ if $n_p$ is infinite. So $\Omega_a^m$ has torsion iff some $n_p$ is finite and nonzero, i.e. if there is a prime that divides some $a_n$ but does not divide infinitely many $a_n$s (for $n\geq m$). This means $\Omega_a$ has torsion iff there is a prime that divides some $a_n$ but does not divide $a_n$s for arbitrarily large $n$.