I believe that the answer to your question "Can $f$ be even and odd at the same time?" is no, but the argument that I have seems more complicated than I expected it to be. The idea is synthetically to imitate the notion of a 'local orientation' and show that it has global coherence. Here is an outline:
Let $H$ be a Hilbert plane and let $G$ be its group of isometries. For each point $P$ in $H$ let $G_P\subset G$ be the subgroup consisting of those isometries of $H$ that fix $P$.
Let $R_P$ denote the set of rays in $H$ that emanate from $P$. Then $G_P$ acts transitively on $R_P$, and the stabilizer in $G_P$ of a ray $r \in R_P$ is a $2$-element subgroup consisting of the identity and the reflection in the line that is the union of $r$ and its opposite ray $-r$.
Now, let $C_P \subset G_P$ denote the set of elements that are either the identity or that fix no ray in $R_P$. Then $C_P$ is a abelian subgroup of index $2$ in $G_P$, and its complement in $G_P$ is exactly the set of reflections in lines through $P$. (It's easy to see that every element of $C_P$ is (not uniquely) the product of two reflections in lines through $P$.)
Now, for any two points $P$ and $Q$ in $H$, there is a canonical identification of $R_P$ with $R_Q$. Namely, if $P=Q$, we take the identification to be the identity. If $P\not=Q$, then consider the line $PQ$ and let $r\in R_P$ denote the ray emanating from $P$ that passes through $Q$. This ray $r$ contains a unique ray $r'\in R_Q$ and we let $(-r)'=-r'\in R_Q$ denote the ray that corresponds to the opposite ray $-r\in R_P$. For any other ray $s\in R_P$, we let $s'\in R_Q$ denote the ray that is on the same side of the line PQ as $s$ and for which the angle made by $r'$ and $s'$ is congruent to the angle made by $r$ and $s$. Call this mapping $B_Q^P:R_P\to R_Q$. Then $B_Q^P$ is a bijection and $B_P^Q$ is its inverse.
If $O$, $P$, and $Q$ are any three points, then the composition $B_P^OB_Q^PB_O^Q:R_O\to R_O$ is the action of an element of $C_O$ on $R_O$. (It is the identity if $O$, $P$, and $Q$ are collinear.)
Meanwhile, if $\rho:H\to H$ is a reflection in a line, then $\rho$ induces a map on rays $\rho'_P:R_P\to R_{\rho(P)}$, and $B^{\rho(P)}_P\circ\rho'_P:R_P\to R_P$ is induced by an element of $G_P$ that is not in $C_P$.
Finally, one can now divide the elements of $G$ into two classes: The even elements $f$ are the isometries such that $B^{f(P)}_P\circ f'_P:R_P\to R_P$
is induced by an element of $C_P$. (This condition turns out to be independent of $P$.) These form a subgroup $G_0\subset G$. The odd elements, which contain all the reflections in lines, are what is left. Moreover, it now follows that $G_0$ consists of products of an even number of line reflections, while its complement consists of all of the products of an odd number of line reflections.
The proofs of all the claims made in this outline (and they do require proof) are straightforward, though tedious.
