Here is what I could figure out based on the elliptic curve notes here: https://crypto.stanford.edu/pbc/notes/elliptic/explicit.html, given a point $P=(x_1,y_1)$ and a curve defined by
$$Y^2+a_1XY = X^3+a_2X^2+a_4X+a_6$$
then the $x$-coordinate of $2P$ is given by
$$x_2 = \lambda^2 + a_1\lambda - a_2 - 2x_1$$
In your case, $a_1=1$. Also $2x_1=0$ because the field has characteristic 2, and we can switch all minus signs to plus signs for the same reason. Thus, the formula becomes:
$$x_2=\lambda^2 + \lambda + a$$
The trace of $x_2$ will be $\text{tr}(\lambda^2) + \text{tr}(\lambda)+\text{tr}(a)$. Since $\text{tr}(\lambda^2) = \text{tr}(\lambda)$, this means $\text{tr}(x)=\text{tr}(a)$. Thus, for any point $P$ on the curve, if $P=2Q$ for some $Q$, then the trace of $P$'s $x$-coordinate equals the trace of $a$.
If we have a cyclic subgroup with odd order $n$, then $2$ has some inverse $2^{-1}$ modulo $n$. Thus, starting from any point $P$ in this subgroup, we know that $2(2^{-1}P)=P$, so there exists a point $Q=2^{-1}P$ such that $P=2Q$, and thus its $x$-coordinate has the same trace as $a$.
Generally, every element of the subgroup $2E(GF(2^m)) = \{x+x : x\in E(GF(2^m))\}$ will all have this same trace.
What about other points? I don't know. It might be that points $P$ which do not equal $2Q$ for any $Q$ can still have trace equal to $\text{tr}(a)$, or maybe you can prove that they cannot.
