I instead tried to construct $(ab)^6+(ab)^4+(ab)^3-(ab)^2-1=0$ using $\,\ldots$
Following up on OP's approach, let $\,ab=p\,$ and, for symmetry, $\,a+b=s\,$. To begin with:
$$
\begin{cases}
\begin{align}
a^4 + a^3 - 1 &= 0 \\
b^4 + b^3 - 1 &= 0
\end{align}
\end{cases} \tag{1}
$$
Subtracting the two equations in $\,(1)\,$, and using that $\,a \ne b\,$:
$$\require{cancel}
\begin{align}
a^4-b^4+a^3-b^3 = 0 \;\;&\implies\;\; \cancel{(a-b)}(a+b)(a^2+b^2) + \cancel{(a-b)}(a^2+ab+b^2) = 0 \\
&\implies\;\; s(s^2-2p) + s^2 - p = 0 \\
&\implies\;\; s^3 + s^2 - (2s+1)p = 0 \tag{2}
\end{align}
$$
Adding the two equations in $\,(1)\,$:
$$
\begin{align}
a^4+b^4+a^3+b^3 - 2 = 0 \;\;&\implies\;\; \big((a+b)^4 - 4ab(a^2+b^2)-6a^2b^2\big) \\
&\quad\quad\quad\quad+\big((a+b)^3-3ab(a+b)\big)-2=0 \\
&\implies\;\; s^4 - 4p(s^2-2p)-6p^2+s^3-3ps-2 = 0 \\
&\implies\;\; s^4 + s^3 -4ps^2-3ps+2p^2-2=0 \tag{3}
\end{align}
$$
At this point, the equation satisfied by $\,p\,$ can be derived by eliminating $\,s\,$ between $\,(2)\,$ and $\,(3)\,$. This is usually better left to a CAS since the calculations can be tedious, and e.g. WA gives resultant[$s^3 + s^2 - (2s+1)p$, $s^4 + s^3 - 4ps^2 - 3ps + 2p^2 - 2, s$] = $8 (p^6 + p^4 + p^3 - p^2 - 1)$.
It can still be done by hand, though. Substituting $\,s^4+s^3=s \cdot (2s+1)p\,$ from $\,(2)\,$ into $\,(3)\,$:
$$
\begin{align}
s(2s+1)p -4ps^2-3ps+2p^2-2=0 \;\;&\implies\;\; -2ps^2 - 2ps+ 2p^2 - 2 = 0 \\
&\implies\;\; ps^2+ps-p^2+1 = 0 \tag{4}
\end{align}
$$
Eliminating once again $\,s(s+1)\,$ between $\,(2)\,$ and $\,(4)\,$ by $\,s \cdot (4) - p \cdot (2)\,$ gives:
$$
\begin{align}
0 &= s \cdot \big(\cancel{ps(s+1)}-p^2+1\big) - p \cdot \big(\cancel{s^2(s+1)} - (2s+1)p\big) = (p^2+1)s+p^2 \tag{5}
\end{align}
$$
Then, substituting $\,s = \frac{-p^2}{p^2+1}\,$ back in $\,(4)\,$ gives in the end the sextic $\,p^6 + p^4 + p^3 - p^2 - 1 = 0\,$.
