Having thought about this again, I think the following works.
(Warning: The argument below seems almost a bit too much on the side of generalities, so maybe someone with more experience can check whether there's not an issue hiding somewhere in the more sloppy parts of it.)
First, let $Y\in \mathsf{CGWH}$ and $K\in \mathsf{CompHaus}$. Then for all $L\in \mathsf{CompHaus}$:
\begin{align*}
\underline{Y}^{\underline{K}}(L) &\cong \hom(\underline{L}, \underline{Y}^{\underline{K}}) \\
&\cong \hom(\underline{L} \times \underline{K}, \underline{Y})\\
&\cong \hom(\underline{L \times K}, \underline{Y}) \\
&\cong \hom(L\times K, Y) \\
&\cong \hom(L, Y^{K}) \\
&\cong \hom(\underline{L}, \underline{Y^{K}})\\
&\cong \underline{Y^{K}}(L).
\end{align*}
Hence, by the Yoneda lemma,
$$ \underline{Y}^{\underline{K}} \cong \underline{Y^{K}}$$
Now, every $X\in \mathsf{Cond}(\mathsf{Set})$ is the colimit of representables, written with an abuse of notation as follows,
$$ X \cong \mathrm{colim}_{K\to X} \underline{K}. $$
Therefore, if $Y\in \mathsf{CGWH}$ and $X\in \mathsf{Cond}(\mathsf{Set})$, then
\begin{align*}
\underline{Y}^X &\cong \lim_{K\to X} \underline{Y}^{y(K)}\\
&\cong \lim_{K\to X} \underline{Y^{K}} \\
&\cong \underline{\lim_{K\to X} Y^{K}} \\
&\cong \underline{Y^{\mathrm{colim}_{K\to X} K}} \\
&\cong \underline{Y^{X(*)_{\mathrm{top}}}},
\end{align*}
where
\begin{align*}
X \mapsto X(*)_{\mathrm{top}} := \mathrm{colim}_{K\to X} K,
\end{align*}
is the left adjoint to the inclusion
$$ \underline{\cdot}: \mathsf{CGWH}\hookrightarrow \mathsf{Cond}(\mathsf{Set}), $$
and the colimit is taken in $\mathsf{CGWH}$. Therefore,
$$ \underline{Y}^X\cong \underline{Y^{X(*)_{\mathrm{top}}}}$$
and $\mathsf{CGWH}$ is indeed an exponential ideal in $\mathsf{Cond}(\mathsf{Set})$.
So it seems to make no difference whether I take my loop space $M^{\mathbb{S}}$, (or space of continuous functions $C(X)$, or whatever) in the more "classical" $\mathsf{CGWH}$ or in condensed sets, as long as the codomain is a CGWH space to begin with.