I am trying to prove the following statement. I don't have any assumptions on the the base field $\Bbbk$ yet, but I am happy to restrict this if required.
Statement: Let $p(h)\in\Bbbk[h^{\pm 1}]$. If $p(h)+p(h^{-1})=m(h)(h^{-1}-h),$ then $p(h)=g(h)(h^{-1}-h)$, where $g(h)\in\Bbbk[h^{\pm 1}]$.
In other words, if $h^{-1}-h$ is a factor of $p(h)+p(h^{-1})$, then $h^{-1}-h$ is a factor of $p(h)$.
For example, let $m(h)=(h^{-1}-h)$ then $m(h)(h^{-1}-h)=h^2-2+h^{-2}$. Rearranging this we get $h^2-1+h^{-2}-1$. So $p(h)=h^2-1=-h(h^{-1}-h).$
Notes
The equation $p(h)+p(h^{-1})=m(h)(h^{-1}-h)$ implies that $m(h)=-m(h^{-1})$. To see this, apply the map $h\mapsto h^{-1}$.
If $p(h)=g(h)(h^{-1}-h)$, then $$g(h)(h^{-1}-h)+g(h^{-1})(h-h^{-1})=\left[g(h)-g(h^{-1})\right](h^{-1}-h).$$
The statement is true (there might be some issues with fields of characteristic $2$, but the statement will be true for all other fields). Let $\mathbb{k}$ be a field of characteristic not equal to $2$. Let $p \in \mathbb{k}[X,X^{-1}]$ be an arbitrary Laurent polynomial, say $$p(X) = \sum_{k=-d}^d a_kX^k.$$ Let us write $$P(X) = p(X) + p(X^{-1}) = \sum_{k=-d}^d (a_k + a_{-k})X^k.$$ The condition $$P(X) = p(X) + p(X^{-1}) = m(X)(X-X^{-1})$$ implies that $P(1) = 0$ and $P(-1) = 0$. Equivalently $$\sum_{k=-d}^d(a_k + a_{-k}) = 2\sum_{k=-d}^da_k= 0,\qquad \text{and} \qquad \sum_{k=-d}^d(-1)^k(a_k + a_{-k}) = 2\sum_{k=-d}^d(-1)^ka_k= 0.$$ This is true if and only if $p(1) = p(-1) = 0$, which is true if and only if $p(X) = g(X)(X-X^{-1})$ for some $g \in \mathbb{k}[X,X^{-1}]$. To see why the latter statement is true, it's easier to pass to (regular) polynomials. Let $\tilde{p}$ be the associated polynomial given by $\tilde{p}(X) = X^dp(X)$. Then $p(1) = p(-1) = 0$ also implies that $\tilde{p}(1) = \tilde{p}(-1) = 0$. The polynomial remainder theorem then implies that $(X^2 -1)\mid \tilde{p}(X)$, i.e., there exists some polynomial $\tilde{g}(X)$ such that $$\tilde{p}(X) = \tilde{g}(X)(X^2-1).$$ Dividing through by $X^d$ gives us $$p(X) = g(X)(X-X^{-1}),$$ where $g(X) = X^{-(d-1)}\tilde{g}(X)$.