$B$ is actually isomorphic to $\mathbb{Z}\oplus\mathbb{Z}$. The general solution to the system
$\begin{cases} 5m+7n=a\\ m+2n=b \end{cases}$
is
$\begin{cases}m=\frac13(2a-7b)\\\\ n=\frac13(5b-a)\;. \end{cases}$
Thus, $\langle a,2b\rangle\in B$ if and only if $3\mid 2a-7b$ and $3\mid 5b-a$, i.e., if and only if $a\equiv 5b\pmod 3$ and $2a\equiv 7b\pmod 3$. These reduce to $a\equiv 2b\pmod 3$ and $b\equiv 2a\pmod 3$, respectively, which are equivalent to each other and to $a+b\equiv 0\pmod 3$. In short,
$\begin{align*}B&=\{\langle a,2b\rangle\in\mathbb{Z}\oplus\mathbb{Z}:3\mid a+b\}\\ &=\{\langle a,b\rangle\in\mathbb{Z}\oplus\mathbb{Z}:a\equiv b\!\!\!\!\pmod 3\text{ and }2\mid b\}\;. \end{align*}$
Thus, $\langle a,b\rangle\in B$ iff there are $m,n\in\mathbb{Z}$ such that $b=2n$ and $a=3m+2n$, and conversely, any $m,n\in\mathbb{Z}$ uniquely determine a pair $\langle 3m+2n,2n\rangle\in B$. Thus, $B=\{m\langle 3,0\rangle+m\langle 2,2\rangle:m,n\in\mathbb{Z}\}\;,$
and the map
$h:\mathbb{Z}\oplus\mathbb{Z}\to B:\langle m,n\rangle\mapsto\langle 3m+2n,2n\rangle$
is an isomorphism.
Now let $\langle m,n\rangle\in A$; then exactly one of the following pairs belongs to $B$:
$\begin{cases}\langle m,n\rangle,\langle m,n-2\rangle,\text{ or }\langle m,n-4\rangle,&\text{if }n\text{ is even}\\ \langle m,n-1\rangle,\langle m,n-3\rangle,\text{ or }\langle m,n-5\rangle,&\text{if }n\text{ is odd}\;. \end{cases}$
Equivalently, every element of $A$ belongs to $\langle 0,k\rangle+B$ for some $k\in\{0,1,2,3,4,5\}$, and from that you should easily be able to answer the question in the title.
It might help to look at a graphical display of part of $A$; members of $B$ are colored red or blue, the two blue elements being the generators for $h$.
$\begin{array}{c} \langle -3,4\rangle&\color{red}{\langle -2,4\rangle}&\langle -1,4\rangle&\langle 0,4\rangle&\color{red}{\langle 1,4\rangle}&\langle 2,4\rangle&\langle 3,4\rangle&\color{red}{\langle 4,4\rangle}&\langle 5,4\rangle\\ \langle -3,3\rangle&\langle -2,3\rangle&\langle -1,3\rangle&\langle 0,3\rangle&\langle 1,3\rangle&\langle 2,3\rangle&\langle 3,3\rangle&\langle 4,3\rangle&\langle 5,3\rangle\\ \langle -3,2\rangle&\langle -2,2\rangle&\color{red}{\langle -1,2\rangle}&\langle 0,2\rangle&\langle 1,2\rangle&\color{blue}{\langle 2,2\rangle}&\langle 3,2\rangle&\langle 4,2\rangle&\color{red}{\langle 5,2\rangle}\\ \langle -3,1\rangle&\langle -2,1\rangle&\langle -1,1\rangle&\langle 0,1\rangle&\langle 1,1\rangle&\langle 2,1\rangle&\langle 3,1\rangle&\langle 4,1\rangle&\langle 5,1\rangle\\ \color{red}{\langle -3,0\rangle}&\langle -2,0\rangle&\langle -1,0\rangle&\color{red}{\langle 0,0\rangle}&\langle 1,0\rangle&\langle 2,0\rangle&\color{blue}{\langle 3,0\rangle}&\langle 4,0\rangle&\langle 5,0\rangle\\ \langle -3,-1\rangle&\langle -2,-1\rangle&\langle -1,-1\rangle&\langle 0,-1\rangle&\langle 1,-1\rangle&\langle 2,-1\rangle&\langle 3,-1\rangle&\langle 4,-1\rangle&\langle 5,-1\rangle\\ \langle -3,-2\rangle&\color{red}{\langle -2,-2\rangle}&\langle -1,-2\rangle&\langle 0,-2\rangle&\color{red}{\langle 1,-2\rangle}&\langle 2,-2\rangle&\langle 3,-2\rangle&\color{red}{\langle 4,-2\rangle}&\langle 5,-2\rangle\\ \langle -3,-3\rangle&\langle -2,-3\rangle&\langle -1,-3\rangle&\langle 0,-3\rangle&\langle 1,-3\rangle&\langle 2,-3\rangle&\langle 3,-3\rangle&\langle 4,-3\rangle&\langle 5,-3\rangle\\ \langle -3,-4\rangle&\langle -2,-4\rangle&\color{red}{\langle -1,-4\rangle}&\langle 0,-4\rangle&\langle 1,-4\rangle&\color{red}{\langle 2,-4\rangle}&\langle 3,-4\rangle&\langle 4,-4\rangle&\color{red}{\langle 5,-4\rangle}\\ \langle -3,-5\rangle&\langle -2,-5\rangle&\langle -1,-5\rangle&\langle 0,-5\rangle&\langle 1,-5\rangle&\langle 2,-5\rangle&\langle 3,-5\rangle&\langle 4,-5\rangle&\langle 5,-5\rangle\\ \color{red}{\langle -3,-6\rangle}&\langle -2,-6\rangle&\langle -1,-6\rangle&\color{red}{\langle 0,-6\rangle}&\langle 1,-6\rangle&\langle 2,-6\rangle&\color{red}{\langle 3,-6\rangle}&\langle 4,-6\rangle&\langle 5,-6\rangle\\ \end{array}$