So there is at least a somewhat natural way to define a composition of ordinals. I describe it briefly here. The proof that everything works is tedious.
First define $\mathbf{\Omega}_0=\omega^{\mathbf{Ord}}$, and for each ordinal $0<\alpha$, define the class $\mathbf{\Omega}_{\alpha}$ as the class of common fixed points of the functions $\Omega_{\beta},\beta<\alpha$ where $\Omega_{\beta}$ is the isomorphism $(\mathbf{Ord},<) \rightarrow (\mathbf{\Omega}_{\beta},<)$.
For $0<\alpha \in \mathbf{Ord}$, let $F_{\alpha}$ denote the isomorphism $(\mathbf{Ord}^{\geq \omega},<) \rightarrow (\mathbf{\Omega}_{\alpha} \setminus \mathbf{\Omega}_{\alpha+1},<)$. The function $F_0$ is defined by $F_0(\omega\otimes(1\oplus \beta)\oplus n)=2^n \omega^{\omega \otimes(1\oplus f(\beta))}$ for all $\beta \in \mathbf{Ord}$ and $n<\omega$, where $\oplus$ and $\otimes$ are the ordinal sum and product and $f$ is the isomorphism $(\mathbf{Ord},<) \rightarrow (\mathbf{Ord} \setminus \mathbf{\Omega}_1,<)$. All functions $F_{\alpha}$ are strictly increasing with $\forall \alpha<\beta \in \mathbf{Ord},\forall \xi \in \mathbf{Ord}^{\geq \omega}, \xi<F_{\alpha}(\xi)<F_{\beta}(\xi)$.
We also have $F_0(\xi+\psi)=F_0(\xi) F_0(\psi)$ and $F_0(\xi+n)=2^nF_0(\xi)$ for all ordinals $\xi,\psi \geq \omega$ and $n<\omega$.
Every ordinal $\alpha$ can be uniquely written as $\alpha=P_0(\omega) \alpha_0 +...+P_r(\omega) \alpha_r$ where $r \in \mathbb{N}$, $\alpha_0>...>\alpha_r$ are ordinals of the form $\alpha_i=\omega^{\beta_i}$ where $\beta_i$ is a limit ordinal, and $P_0,...,P_r \in \mathbb{N}[X]$.
In this notation, each ordinal $\alpha_i$ for $0 \leq i \leq r$ with $\alpha_i>1$ can be written uniquely as $F_{\nu_{\alpha_i}}(\alpha_i')$ for certain $\nu_{\alpha_i} \in \mathbf{Ord}$ and $\alpha_i' \in \mathbf{Ord}^{\geq \omega}$. We define $\alpha \circ \beta$ inductively as $\alpha \circ \beta = \sum \limits_{\alpha_i>1} P_i(\beta)F_{\nu_{\alpha_i}}(\alpha_i' \circ \beta)+ \sum \limits_{\alpha_r=1} P_r(\beta)$.
This makes $\circ$ the unique composition of ordinals such that for $\alpha \in \mathbf{Ord}$ and $\xi \in \mathbf{Ord}^{\geq \omega}$, we have $F_{\alpha}(\omega) \circ \xi = F_{\alpha}(\xi)$.
For the general method to work, the function $F_0$ should transform sums into products and the ranges of the other functions $F_{\alpha}$ should be mutually disjoint subclasses of the class of ordinals $\omega^{\beta}$ where $0<\beta$ is limit. But there still is some freedom in the choice of functions $F_{\alpha}$. For instance one can make it so each function $F_{\alpha+1}$ is a solution of the Abel equation $\forall \omega\leq\xi,F_{\alpha+1}(\xi+1)=F_{\alpha}(F_{\alpha+1}(\xi))$.