Is it possible to make this proof two-way? 
Show that if $H_1$ and $H_2$ are cyclic groups, their direct product is cyclic if their orders are coprime.

If cyclic, then there exists an element $(E_1 | E_2)$ such that $(E_1 | E_2)^n=(h_1|h_2)$ for any combination of $h_1,h_2$ for some $n\in \mathbb N$.
Because of the definition of the operation on the new group we have that this is exactly the case when $E_1^n=h_1 \land E_2^n=h_2$.
Now, somewhere in the following lines the equivalence probably gets lost (and maybe the sufficiency, too):
It follows that for some $m\in \mathbb N$, $E_1^m=E_1E_1^n \land E_2^m=E_2^n$, thus (because for a generator $g$ of $G$ $g^n=g^{n \mod|G|}$)
$$m=1+n\mod|H_1|$$$$m=n \mod|H_2|$$
We can write $|H_1|=ad$ and $|H_2|=bd$ where $d=\gcd(|H_1|,|H_2|)$:
$$1+ad=bd$$$$1=d(b-a)$$
So $d$ must be $1$.
Is it possible to make the way up/backwards from the last lines? 
 A: You're tacitly assuming that $H_1$ and $H_2$ are finite. But I'm afraid you get lost much earlier than you think.
Suppose the direct product $H_1\times H_2$ is cyclic. Then both factors are cyclic, because there are surjective homomorphisms $H_1\times H_2\to H_1$ and $H_1\times H_2\to H_2$ (the projections).
We want to show that the two groups have coprime orders. Let $m=|H_1|$ and $n=|H_2|$ and let $d=\gcd(m,n)$. If $x\in H_1$ and $y\in H_2$, then
$$
(x,y)^{mn/d}=((x^{m})^{n/d},(y^{n})^{m/d})=(1,1)
$$
so the order of any element divides $mn/d$. Since an element of order $mn$ exists by hypothesis, we get $d=1$.
Conversely, suppose $\gcd(m,n)=1$ and let us show that $H_1\times H_2$ is cyclic. Let $H_1=\langle u\rangle$ and $H_2=\langle v\rangle$. Consider $x=u^r\in H_1$ and $y=v^s\in H_2$.
By the Chinese remainder theorem, the system of congruences
$$
\begin{cases}
t\equiv r\pmod{m}\\
t\equiv s\pmod{n}
\end{cases}
$$
is solvable. So let $t=r+ma=s+nb$. Then
$$
(u,v)^t=(u^{r+ma},v^{s+nb})=(u^r,v^s)=(x,y)
$$
and therefore $H_1\times H_2=\langle(u,v)\rangle$ is cyclic.

If you don't want to apply the Chinese remainder theorem, you can prove it. Since $H_1\cong \mathbb{Z}/m\mathbb{Z}$ and $H_2\cong\mathbb{Z}/n\mathbb{Z}$ we can just work with the two easier groups.
Consider the map
$$
\varphi\colon\mathbb{Z}\to
\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}
$$
defined by
$$
\varphi(z)=(z+m\mathbb{Z},z+n\mathbb{Z})
$$
It is clear that $\varphi$ is an isomorphism and that
$$
\ker\varphi=m\mathbb{Z}\cap n\mathbb{Z}=k\mathbb{Z}
$$
because $k$ is the lowest common multiple of $m$ and $n$ (prove the last equality). Since $\gcd(m,n)=1$, we have $k=mn$. Thus we have an injective homomorphism
$$
\bar\varphi\colon\mathbb{Z}/(mn)\mathbb{Z}\to
\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}
$$
which, by comparing the cardinalities, is surjective. So $\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}$ is isomorphic to $\mathbb{Z}/mn\mathbb{Z}$ and therefore it is cyclic.
