Not an answer yet but too long for a comment.

For the $a_n$ as in მამუკა ჯიბლაძე's answer, it is worth to look at even and odd indices separately. In fact, for $n$ odd, $a_n$ only seems to have terms for $x^{n-4k}$ and $[x^{n-4k}]a_n=[x^{n-4k-1}]a_{n-1}$, so all the information is contained in the even polynomials.
Writing $r$ for $\varepsilon_{2^r}$ and keeping the signs, we have the following coefficients:

$\begin{array} {rrrrrrrrrrrrrrrr} n&|&x^{n-2}&x^{n-4}&.&x^{n-8}&.&.&.&x^{n-16}\\ \hline 2&|&1\\ \hline 4&|&2&12\\ 6&|&-1&&-2\\ \hline 8&|& 3&-13&&-23\\ 10&|& 1&&-3&-23&-123\\ 12&|& -2&-12&&&3&13\\ 14&|& -1&&2&&&&-3\\ \hline 16&|& 4&-14&&24&&&&-34\\ 18&|& 1&&-4&24&124&&&-34&-134\\ 20&|& 2&12&&&4&14&&-34&-234&-1234\\ 22&|& -1&&-2&&&&-4&-34&134&&234\\ \end{array}$

Except for the signs, which keep some mystery, the patterns of the columns seem already essentially predictable, everything in terms if the A006519 sequence. If you compute some more lines, it should be entirely clear (e.g. what happens in the 5th column?), and from there of course possible to prove that by some induction.

Also note that the "main diagonal" of this table, taken as a binary code, seems to encapsulate a bijection $ 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, ...$ from $\mathbb N\to\mathbb N$, more precisely the Gray code A003188.

Not an answer yet but too long for a comment.

For the $a_n$ as in მამუკა ჯიბლაძე's answer, it is worth to look at even and odd indices separately. In fact, for $n$ odd, $a_n$ only seems to have terms for $x^{n-4k}$ and $[x^{n-4k}]a_n=[x^{n-4k-1}]a_{n-1}$, so all the information is contained in the even polynomials.

Writing $r$ for $\varepsilon_{2^r}$ and keeping the signs, we have the following coefficients:

$\begin{array} {rrrrrrrrrrrrrrrr} n&|&x^{n-2}&x^{n-4}&.&x^{n-8}&.&.&.&x^{n-16}\\ \hline 2&|&1\\ \hline 4&|&2&12\\ 6&|&-1&&-2\\ \hline 8&|& 3&-13&&-23\\ 10&|& 1&&-3&-23&-123\\ 12&|& -2&-12&&&3&13\\ 14&|& -1&&2&&&&-3\\ \hline 16&|& 4&-14&&24&&&&-34\\ 18&|& 1&&-4&24&124&&&-34&-134\\ 20&|& 2&12&&&4&14&&-34&-234&-1234\\ 22&|& -1&&-2&&&&-4&-34&134&&234\\ \end{array}$

Except for the signs, which keep some mystery, the patterns of the columns seem already essentially predictable, everything in terms if the A006519 sequence. If you compute some more lines, it should be entirely clear (e.g. what happens in the 5th column?), and from there of course possible to prove that by some induction.

If one adds to your list of $\varepsilon_{n}$'s the lines $\varepsilon_{2^k}=\varepsilon_{2^k}$ for completeness, then the pattern of indices 2,2,4,4,2,2,8,8,2,2,4,4,2,2,16,16... in particular also should be isomorphic to the (doubled) A006519 sequence. But the signs...? (Sadly, $\varepsilon_{22}=-\varepsilon_{2}$.)

Also note that the "main diagonal" of this table (i.e. the array of the absolute terms of the even $a_n$), taken as a binary code, seems to encapsulate a bijection $ 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, ...$ from $\mathbb N\to\mathbb N$, more precisely the Gray code A003188.