Let $E = \{2k: k \in \Bbb{N}\}$, and let $M = \{m = (2r)(4a + 2) : r, a \in \Bbb{N}\}$.
Prove that some elements in $E$ are not uniquely representable as products of elements of $M$, e.g. $2(30) = (2(3))(4(2) + 2) = (2(1))(4(7) + 2)$.
We have to show that $e_1 \in E$ is representable as two products of even primes, so $e = (2r_1)(4a_1 + 2) = (2r_2)(4a_2 + 2)$ and so we end up with $$2(r_1a_1 - r_2a_2) = r_1 - r_2$$
Now we know the trivial solution is $r_1 = r_2 = 0$ but then $e = 0$. So we suppose $r_1 \neq r_2$. Then, $r_1 - r_2 = \text{even}$ necessarily; either $r_1, r_2$ are both odd or both even. But I do not know how to continue.
Basically the problem can be translated as: prove some number $k$ can be written in more than one way as $m\cdot n$ with $m$ even and $n$ a multiple of $2$ but not of $4$.
Let $k=2^rd$ with $d$ odd. Notice that if $r<2$ there are zero representations of $k$. So we assume $r\geq 2$.
We prove that $k=2^rd$ has a unique representation as $m\cdot n$ with $m$ even and $n$ multiple of $2$ but not of $4$ if and only if $d=1$ or $r=2$ and $d$ is prime.
To see why notice that $n\cdot m=k$ satisfy the requirements if and only if $n=2^{r-1}{d'}$ with $d'|d$. So the number of representations is equal to the number of divisors of $d$, except in the case where $r=2$, in this case the number of representations is equal to the number of divisors of $d$ that are less than or equal to $d'$ (this is because both $m$ and $n$ are of the form $2*d'$ with $d'$ a divisor of $d$.