The set $H := \bigoplus\limits_{j=-\infty}^{\infty} G$ forms a group under componentwise group law on $G$. That is the direct product and that means essentially the same thing as $H' = \prod\limits_{j={-\infty}}^{\infty} G = \dots \times G \times \dots$ except there is only a finite number of non-identity entries in each element. In either case, component-wise appylying $G$'s group law would form a group in these bi-infinite products.
Clearly we can write componentwise $\cdot$ as $x \cdot y = (x_i \cdot y_i)$ where $x, y \in H$. Then correspondingly we can write component-wise "shift second argument by $k$, and then perform $\cdot$" as $x \cdot_k y = (x_i\cdot y_{i-k})$ so was wondering if a group was also formed or if not, then is it at least associative and thus a monoid with identity $1_H = (\dots, 1, \dots)$ for any $k \in \Bbb{Z}$?
What's the simplest proof you can give of associativity of $\cdot_k$, knowing that $\cdot \equiv \cdot_0$ is associative by definition of group $G$ and component-wise constructions?
If you need to, choose a basepoint, such as $0$, which would mean that $(a,b) := (\dots, 1, a, b, 1, \dots)$ has an $a$ at index $0$ and a $b$ at index $1$, and so on.
The new operation is not associative. But in a way it fits into another associative operation, if we allow $k$ to vary and keep track of its value.
The mapping $$\phi:(y_i)\mapsto(y_{i-k})$$ is an automorphism of $H$. Your product is defined as $$ x\cdot_k y=x\cdot \phi(y). $$ As you observed, this is not associative, because $$ (x\cdot_k y)\cdot_kz=(x\cdot\phi(y))\cdot\phi(z), $$ but $$ x\cdot_k(y\cdot_kz)=x\cdot(\phi(y)\cdot\phi(\phi(z))), $$ and in the latter product $\phi$ is applied to $z$ twice.
But this is related to the concept of a semidirect product of groups. Whenever we have a homomorphism $\phi$ from a group $G$ to the group of automorphisms of another group $H$, we can form the semidirect product $H\rtimes_\phi G$. The underlying set is the usual Cartesian product $H\times G$, but the operation defined by the rule $$ (h,g)*(h',g')=(h\phi(g)(h'),gg'). $$ In other words, we twist the $H$-component of the latter factor by applying the automorphism $\phi(g)$ to it before we multiply.
The rule $\phi(k):(x_i)\mapsto (x_{i-k})$ defines a homomorphism from the additive group $\Bbb{Z}$ to the group of automorphisms of $H$. In other words, for all $k\in\Bbb{Z}$, the shift mapping $\phi(k)$ is an automorphism of $G$, and furthermore $$\phi(k)\circ\phi(\ell)=\phi(k+\ell)$$ for all integers $k,\ell$. This is related to your question as follows. In the semi-direct product $H\rtimes_\phi\Bbb{Z}$ we have $$ (x,k)*(y,\ell)=(x\cdot_ky,k+\ell). $$ You see that the group operation of the semidirect product keeps track of the "combined" shift (here $k+\ell$), and that makes the operation $*$ associative.