When asked to prove that "$\exists$ a ..."; then what you are really doing is actually finding whatever is that you need to prove exists.
For example in your question you must prove that there exists a one-to-one correspondance between $\mathbb{Z}^{+}$ and positive even integers, which I will now denote $\mathbb{Z}_e$ so we should attempt to find a map which takes $\mathbb{Z}^{+} \rightarrow \mathbb{Z}_e$.
So how should we go about finding one? well first lets think what is the formal definition of even? I would say an integer $x$ is even if $x = 2k$ for some $k\in \mathbb{Z}$ so the set of positive even integers is $\mathbb{Z}_e = \{x = 2k : k\in\mathbb{Z}^{+}\}$.
Now once we have actually formalized what a positive even integer is it is not hard to think of a map, for example take:
$f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}_e$ defined by :
$k \mapsto 2k$
Now we've got a map we think we will work, and we just need to check if it is one-to-one.
Suppose $f(r) = f(s)$ Then $2r = 2s$, but this quickly implies that $r = s$ so the map is one-to-one, as desired.
Furthermore the map is also onto, because $\mathbb{Z}_e = \{x = 2k : k\in \mathbb{Z}^{+}\}$ is the set of integers of the form $2k$ by definition.
For some fixed $a \in G$ the map $aH \to Ha, ah \mapsto ha$ is bijective, yes. But this only deals with two specific cosets. It is a completely different statement that there is a $1:1$ correspondence between the set of all left cosets $G ~/~ H$ and the set of all right cosets $H \setminus G$. Observe that the map $G ~/~ H \to H \setminus G$, $aH \mapsto Ha$ is not well defined. In fact, we have $aH = bH \Leftrightarrow (aH)^{-1} = (bH)^{-1} \Leftrightarrow H a^{-1} = H b^{-1}$. So instead, we have to take $G ~/~ H \to H \setminus G$, $aH \mapsto H a^{-1}$. This is a well-defined bijection.
Best Answer
As noted above, you must find a bijection $f$ between $G/H$ (the set of left cosets) and $H\backslash G$ (the set of right cosets).
The standard bijection is such that $f(gH) = Hg^{-1}$. The first step is to show that such a mapping indeed exists, that is that whenever $gH = g'H$ then $Hg^{-1} = Hg'^{-1}$. Thus you can define $f(X)$ as $Hg^{-1}$ for any $g \in X$ without the result depending on the choice of such a $g$.
This step is essential. Note that in it lies the reason for not defining $f$ simply by $f(gH) = Hg$; if $gH = g'H$, it is not generally the case that $Hg = Hg'$.
Then you must show that $f$ is onto and that it is one to one.