$\newcommand\R{\mathbb R}\newcommand{\N}{\mathbb N}\newcommand\la\lambda\newcommand{\be}{\beta}\newcommand{\al}{\alpha}\newcommand\La\Lambda$It is given that the random variable (r.v.) $\La$ has the gamma distribution with parameters $\al,m$. Let $\be:=m$ -- this is being done so because we want to denote by $m$ values of the random variable (r.v.) $M$.
Let $U_1$ and $U_2$ be independent r.v.'s uniformly distributed on $(0,1)$. For positive real $\la,\al,\be$, let $F_\la$ and $F_{\al,\be}$ denote, respectively, the c.d.f.'s of the Poisson distribution with parameter $\la$ and the gamma distribution with parameters $\al,\be$. Then $F_\la^{-1}(U_1)$ and $F_{\al,\be}^{-1}(U_2)$ will be independent r.v.'s with the Poisson distribution with parameter $\la$ and the gamma distribution with parameters $\al,\be$, respectively; as usual, for any c.d.f. $F$ and any $u\in(0,1)$, we let
\begin{equation*}
F^{-1}(u):=\min\{u\in\R\colon F(u)\ge u\}.
\end{equation*}
Then, your formula (1) will hold if $\La=F_{\al,\be}^{-1}(U_2)$ and $N=h(U_1,U_2)$, where
\begin{equation*}
h(u_1,u_2):=F_{F_{\al,\be}^{-1}(u_2)}^{-1}(u_1)
\end{equation*}
for $u_1,u_2$ in $(0,1)$. Informally, given $U_2=u_2$, we first generate the value $\la:=F_{\al,\be}^{-1}(u_2)$ of $\La$ and then, given $U_1=u_1$, the value $n:=F_\la^{-1}(u_1)$ of $N$. Without loss of generality (wlog), for $j\in\{1,2\}$,
\begin{equation*}
U_j=\sum_{i=1}^\infty\frac{B_{ij}}{2^i},
\end{equation*}
where the $B_{ij}$'s are independent Bernoulli r.v.'s with parameter $1/2$.
So, wlog
\begin{equation*}
N=f(B_{11},B_{12},B_{21},B_{22},B_{31},B_{32},\dots) \tag{10}\label{10}
\end{equation*}
and
\begin{equation*}
\La=l(B_{11},B_{12},B_{21},B_{22},B_{31},B_{32},\dots) \tag{20}\label{20}
\end{equation*}
for certain Borel functions $f$ and $l$ on $\R^\N$.
On the other hand, we want to construct r.v.'s $M,X_1,X_2,\dots$, defined on the same probability space as $N$, so that
that
\begin{equation*}
N=g(M,X_1,X_2,\dots), \tag{30}\label{30}
\end{equation*}
where $g(m,x_1,x_2,\dots):=x_1+\dots+x_m$ for nonnegative integers $m,x_1,x_2,\dots$.
The latter task can be carried out as follows. Let $U$ be a r.v. uniformly distributed on $(0,1)$. Let $G_{p,\al}$ be the c.d.f. of the negative binomial distribution with parameters $p,\al$. Forgetting any previous definitions of $N$, let now
\begin{equation*}
N:=G_{p,\al}^{-1}(U), \tag{40}\label{40}
\end{equation*}
so that $N$ has the negative binomial distribution with parameters $p,\al$.
All the r.v.'s in question will be considered defined on the standard probability space $((0,1),|\cdot|)$, where $|\cdot|$ is the Lebesgue measure. In particular, the r.v. $U$ may be assumed defined by the formula $U(u):=u$ for $u\in(0,1)$.
Let
\begin{equation*}
p_{n,m,x_1,\dots,x_m}:=P(N=n,M=m,X_1=x_1,\dots,X_m=x_m), \tag{50}\label{50}
\end{equation*}
with $p_n=P(N=n)$ and $p_{n,m}=P(N=n,M=m)$,
assuming for a second at this point that $N,M,X_1,X_2,\dots$ are as in the paragraph of your post containing formula (3) and $n,m,x_1,x_2,\dots$ are the corresponding values such that $p_{n,m,x_1,\dots,x_m}>0$; of course, actually we have to construct r.v.'s $M,X_1,X_2,\dots$ so that all the equalities \eqref{50} hold -- assuming \eqref{40}.
For $n=0,1,\dots$, consider the disjoint intervals
\begin{equation*}
E_n:=\{u\in(0,1)\colon G_{p,\al}^{-1}(u)=n\}.
\end{equation*}
Then, by \eqref{40}, $N=n$ on $E_n$, and $P(N=n)=|E_n|=p_n$, as desired.
Next, for each $n$, partition $E_n$ into intervals $E_{n,m}$ so that $|E_{n,m}|=p_{n,m}$ for each $m=0,1,\dots$. Letting now $M=m$ on $E_{n,m}$, we get $P(N=n,M=m)=|E_{n,m}|=p_{n,m}$ for all $n,m$, as desired.
Next, partition each interval $E_{n,m}$ into intervals $E_{n,m,x_1}$ so that $|E_{n,m,x_1}|=p_{n,m,x_1}$ for each $x_1=1,2,\dots$. Letting now $X_1=x_1$ on $E_{n,m,x_1}$, we get $P(N=n,M=m,X_1=x_1)=|E_{n,m,x_1}|=p_{n,m,x_1}$ for all $n,m,x_1$, as desired.
Continuing in this manner, we will indeed construct r.v.'s $M,X_1,X_2,\dots$ so that all the equalities \eqref{50} hold -- assuming \eqref{40}. Moreover, then \eqref{30} will hold almost surely (a.s.), because
\begin{equation*}
\begin{aligned}
& P(N\ne g(M,X_1,X_2,\dots)) \\
&=\sum_{n=0}^\infty\sum_{m=0}^\infty\sum_{x_1=1}^\infty\dots\sum_{x_m=1}^\infty
p_{n,m,x_1,\dots,x_m}\,1(n\ne\textstyle\sum_{i=1}^m x_i) \\
&=\sum_{n=0}^\infty\sum_{m=0}^\infty\sum_{x_1=1}^\infty\dots\sum_{x_m=1}^\infty
0=0.
\end{aligned}
\end{equation*}
The above construction used expression \eqref{30} of $N$ as a function of discrete r.v.'s $M,X_1,X_2,\dots$.
Quite similarly, we can construct independent Bernoulli r.v.'s $B_{ij}$ with parameter $1/2$ so that \eqref{10} hold -- for the same r.v. $N$, defined by \eqref{40}. Letting now $\La$ be defined by \eqref{20}, we will satisfy condition (1) in your post. $\quad\Box$
