= Solution
Write $\tau_\varepsilon=T_\varepsilon\wedge T_M$. The stopped <Bessel process> has the same law as the weighted stopped <Brownian motion>, so for every bounded measurable functional $F$ of a continuous path,
$$
\mathbb E F(R^{\tau_\varepsilon})
=\mathbb E_{\mathbb W_x}\left[F(X^{\tau_\varepsilon})
\left(\frac Mx\mathbf1_{\{T_M<T_\varepsilon\}}
+\frac\varepsilon x\mathbf1_{\{T_\varepsilon<T_M\}}\right)\right].
$$
The <Bessel process> hits $M$ in finite time and never hits zero. Its minimum before that hitting time is strictly positive. Thus for sufficiently small $\varepsilon$, depending on the path, its stopped path $R^{\tau_\varepsilon}$ equals $R^{T_M}$ exactly. Under <Wiener measure>, the events $\{T_M<T_\varepsilon\}$ increase to $\{T_M<T_0\}$, and on that event the stopped paths eventually equal $X^{T_M}$. The remaining term is bounded by $\varepsilon\lVert F\rVert_\infty/x$. The <dominated convergence theorem> therefore yields the precise limiting identity
$$
\boxed{\mathbb E F(R^{T_M})
=\frac Mx\mathbb E_{\mathbb W_x}\left[
\mathbf1_{\{T_M<T_0\}}F(X^{T_M})\right].}
$$
The bounded stopped <martingale> $X_{t\wedge T_0\wedge T_M}$ gives
$$
x=\mathbb E X_{T_0\wedge T_M}=M\mathbb W_x(T_M<T_0),
\qquad \mathbb W_x(T_M<T_0)=x/M.
$$
Hence the limiting identity says
$$
\boxed{\mathcal L(R^{T_M})
=\mathcal L_{\mathbb W_x}(X^{T_M}\mid T_M<T_0).}
$$
This establishes <Brownian conditioning by a stopped Bessel density>.
The displayed density in the question requires a stopping convention. On the full, unstopped path space, the probability measure
$$
\frac{d\widetilde{\mathbb Q}^{M}}{d\mathbb W_x}
=\frac Mx\mathbf1_{\{T_M<T_0\}}
$$
is well-defined, but its paths continue as <Brownian motion> after $T_M$. Its pushforward by the stopping map $\omega\mapsto\omega(\cdot\wedge T_M(\omega))$ is the law $\mathbb Q^M$ of the stopped <Bessel process>. Equivalently, if $\mathbb W_x^{0,M}$ denotes the law of <Brownian motion> stopped at $T_0\wedge T_M$, then the actual stopped laws satisfy
$$
\boxed{\frac{d\mathbb Q^M}{d\mathbb W_x^{0,M}}
=\frac Mx\mathbf1_{\{\text{exit at }M\}}.}
$$
Literal absolute continuity of the stopped law with respect to unstopped <Wiener measure> is false: paths that become permanently constant at $M$ have probability one for the stopped law and zero under <Wiener measure>. The same density is also valid on the stopped sigma-algebra of the original coordinate space. Thus the corrected stopped-law formulation proves the intended conclusion without equating these different full-path measures.
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