[step:Identify the same relativized constructible level inside both models]Because $\operatorname{Ord}^{M}=\operatorname{Ord}^{N}$, the ordinal $\alpha$ found in $N$ belongs to $\operatorname{Ord}^{M}$. We claim that $M$ and $N$ compute the same set $L_\alpha[a]$.
This is proved by induction on $\beta \leq \alpha$. At $\beta=0$, both models have the same empty initial level relative to $a$. At successor stages, the defining recursion is
\begin{align*}
L_{\beta+1}[a]=\operatorname{Def}(L_\beta[a],\in,a\cap L_\beta[a]),
\end{align*}
where $\operatorname{Def}(X,\in,A)$ denotes the set of all subsets of $X$ first-order definable over the structure $(X,\in,A)$ from parameters in $X$. Since the induction hypothesis gives the same structure $L_\beta[a]$ in $M$ and $N$, and first-order satisfaction for the set-sized structure $(L_\beta[a],\in,a\cap L_\beta[a])$ is absolute between transitive models containing that structure, both models form the same definable subsets. At limit stages $\lambda \leq \alpha$, both models take the union
\begin{align*}
L_\lambda[a]=\bigcup_{\beta \in \lambda} L_\beta[a],
\end{align*}
and the index set $\lambda$ is the same ordinal in both models.
To justify the successor step more explicitly, fix $\beta<\alpha$ and assume $L_\beta[a]^M=L_\beta[a]^N$. The formulas used in the definability operation are finite strings of natural numbers, so $M$ and $N$ have the same formula codes. The parameters are exactly elements of the common set $L_\beta[a]$. First-order satisfaction for the set-sized structure $(L_\beta[a],\in,a\cap L_\beta[a])$ is absolute between transitive models containing that structure, because its satisfaction relation is obtained by the usual finite recursion on formula complexity. Therefore a subset of $L_\beta[a]$ is definable over this structure from allowed parameters in $M$ if and only if it is definable in $N$. The Separation and Replacement instances included in $\mathrm{ZFC}^{-}_0$ ensure that both models collect exactly these externally characterized definable subsets at stage $\beta+1$, so their internal $\operatorname{Def}$ operations agree.
Hence
\begin{align*}
L_\alpha[a]^M = L_\alpha[a]^N.
\end{align*}[/step]