If I take the HOMFLY(PT) polynomial defined by $$l \,P(L_+) + l^{-1}\,P(L_-) + m\,P(L_0) = 0,$$ I have looked at expressions of the form
(knots that are the same except inside a small disk, where they look like the pictures indicated).
In every case the result had a factor $(l^4 + 2l^2+1-l^2m^2)$.
My question is: why does the expression always have a factor $(l^4 + 2l^2+1-l^2m^2)$?
I understand that this happens when one of the links are disjoint due to the HOMFLY relation relating the disjoint sum (split union) and the connected sum: $$P(L_1 \sqcup L_2)=-\frac{l+l^{-1}}{m} P(L_1 \# L_2),$$ since if you stick this in, you get exactly the factor $(l^4 + 2l^2+1-l^2m^2)$.
Does the relation
perhaps hold in general? According to the proof for the connected sum formula, it shouldn't.
The intuition of the HOMFLY polynomial is that it is the most-general skein relation on $L_+$, $L_-$, and $L_0$. In principle, the HOMFLY polynomial can be given as some relationship between any three distinct compatible rational tangles, so if there were a relation like the one you conjecture at the end, which involves only two tangles, it would necessarily be for some specialization of the HOMFLY polynomial.
Taking your conjectured relation and adding twists to the bottom in two different ways yields the following equality:
That is, if the HOMFLY polynomial satisfied this, then every knot would have the same HOMFLY polynomial. Substituting these back into the HOMFLY skein relation gives the implication that $(\ell+\ell^{-1})^2=m^2$, so indeed it would only be true for some specialization.
For your first observation, that a factor of $\ell^4+2\ell^2+1-\ell^2m^2$ appeared, consider the following. First, it factors as $(1+\ell^2-\ell m)(1+\ell^2+\ell m)$. So, if $m=\pm(\ell+\ell^{-1})$, the resulting polynomial would be zero. Conversely, if the polynomial is zero at both of these evaluations then the polynomial would have the given factor. Let's think about your computation as dealing with diagrams in $S^2$ decomposed as the union of two disks. If each disk contains a tangle, we can think about it as being a bilinear pairing on tangles. By using the HOMFLY skein relation, any given tangle can be reduced to a linear combination of the two crossing-free tangles present in $e$. Hence, we compute the pairing between an arbitrary tangle $T$ and the given element $e$ as follows:
One can check that each of the two terms under the assumption $m=\pm(\ell+\ell^{-1})$ is zero, and hence the pairing of the original tangle with $e$ would be zero. This proves the result.
We can also see what knot invariant you get from this substitution of $m$. The above result gives that $e$ is "equivalent" to $0$ in the sense that plugging it into a diagram yields $0$ (it is in the radical of the bilinear pairing). So, we have the following:
Since split unknots evaluate to $-1$, the HOMFLY polynomial at this evaluation gives $(-1)^{c-1}$, where $c$ is the number of components in the link.