I have $a,b\in{G}$ and also $G$ is Abelian.
I thought that first I should show that $H\subset{G}$ and then show that $\forall{h_1,h_2}\in{H}$ we will have $ h_1\cdot{h_2}\in{H}$.
First to show that $H$ is a subset of $G$, I have taken $a^m\cdot{b^n}\in{H}$. Since $a,b\in{G}$ we have $a^m\cdot{b^n}\in{G}$ from the closure axiom. Secondly to show that $H$ is subgroup I think I need to show first that $H$ is also Abelian but I'm having difficulties approaching this proof.
If I prove that $H$ is Abelian then if I take $h_1=a^m\cdot{b^n},h_2=a^x\cdot{b^y}\in{H}$. Then $$h_1\cdot{h_2}=a^m\cdot{b^n}\cdot{a^x}\cdot{b^y}=a^{m+x\in{\mathbb{Z}}}\cdot{b^{n+y\in{\mathbb{Z}}}}\in{H}.$$
Is this the correct approach? If yes the how can I show that $H$ is Abelian ?
One way to do this is the One-step Subgroup Test.
You have shown that $H\subseteq G$.
We have $e=a^0b^0\in H$, so $H\neq\varnothing$.
Let $g=a^nb^m, h=a^rb^s\in H$ for $n,m,r,s\in\Bbb Z$. Then
$$\begin{align} gh^{-1}&=(a^nb^m)(a^rb^s)^{-1}\\ &=(a^nb^m)(b^{-s}a^{-r})\\ &=a^n(b^m(b^{-s}a^{-r}))\\ &=a^n((b^mb^{-s})a^{-r})\\ &=a^n(b^{m-s}a^{-r})\\ &=a^n(a^{-r}b^{m-s})\quad (G\, (\text{ and hence } H)\text{ is abelian})\\ &=(a^na^{-r})b^{m-s}\\ &=a^{n-r}b^{m-s}, \end{align}$$
but $n-r, m-s\in\Bbb Z$. Hence $gh^{-1}\in H$.
Hence $H\le G$.
(It is not necessary to check whether $H$ is abelian to verify that it is a subgroup. However, since I made use of commutativity, consider that for any $p,q\in H\subseteq G$, we have $pq=qp$ since $G$ is abelian. Hence $H$ is abelian. )