Let $M$ be a non-negative martingale with $\mathbb{E}[M_n] = 1$ for all $n$. Define $\mathbb{Q}_n(F) = \mathbb{E}[\mathbb{1}_F M_n]$ for $F \in \mathcal{F}_n (n \in \mathbb{N})$. I have shown that for all $n \in \mathbb{N}$ and $k \in \mathbb{N}$ one has $\mathbb{Q}_{n+k}(F) = \mathbb{Q}_{n}(F)$ for $F \in \mathcal{F}_n$.
Assume that $M$ is uniformly integrable. Now, I want to show that there exists a probability measure $\mathbb{Q}$ on $\mathcal{F}_\infty = \sigma( \cup_{n \in \mathbb{N}} \mathcal{F}_n)$ that is absolutely continuous w.r.t. $\mathbb{P}$ and that is such that for all $n$ the restriction of $\mathbb{Q}$ to $\mathcal{F}_n$ coincides with $\mathbb{Q}_n$.
When $\mathbb{Q} \ll \mathbb{P}$, I have already shown that $\mathbb{Q}_n \ll \mathbb{P}_n$ and that $\frac{d\mathbb{Q}_n}{d \mathbb{P}_n}=\mathbb{E}_\mathbb{P}[Z | \mathcal{F}_n]$ where $Z = \frac{d\mathbb{Q}}{d \mathbb{P}}$.
However, I am stuck the other way around. Any help is appreciated.
By your assumption there exists a random variable $M_\infty$ such that $M_n \to M_\infty$ $P$-a.s. and in $L^1(P)$ (a.s. because $(M_n)$ is a martingale bounded in $L^1$, and convergence in $L^1(P)$ because of a.s. convergence + uniform integrability).
Now for $F \in {\cal F}_\infty$, define $Q(F) = E_P [ {\bf 1}_F M_\infty]$. Clearly, $Q$ is a probability measure satisfying $Q\ll P$ with $\frac{dQ}{dP} = M_\infty$, $P$-a.s. Also observe that since $(M_n)$ is a $(P,{\cal F}_\cdot)$ martingale, for $F \in {\cal F}_n$ we have
$$ E_P [{\bf 1}_F M_n ] =E_P [ {\bf 1}_F M_{n+k}],~k\ge 0.$$
Since $M_n \to M_\infty$ in $L^1(P)$, it follows that
$$\lim_{k\to\infty} E_P[{\bf 1}_F M_{n+k} ] =E_P[{\bf 1}_F M_\infty].$$
Therefore, from the definition of conditional expectation,
$$M_n = E_P [ M_\infty |{\cal F}_n] .$$
Final note: A martingale $(M_n)$ is called closable (or closed) if there exists an integrable random variable $M_\infty$ such that $E[M_\infty|{\cal F}_n]= M_n$. UI Martingale $\Leftrightarrow$ $L^1$-convergent Martingale $\Leftrightarrow$ Closable Martingale. For this problem we showed the forward implication.