Yes. Here is the strategy that I typically use to derive symmetric-functions
identities from symmetric-polynomials identities. It might not be the most
general strategy, but it has so far been sufficient for myself.

Let $\mathbf{k}$ be the base ring. Consider the $\mathbf{k}$-algebra
$\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ of
formal power series in infinitely many commuting indeterminates $x_{1}
,x_{2},x_{3},\ldots$. If $\mathfrak{m}$ is a monomial and $f\in\mathbf{k}
\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ is a power series,
then $\left[ \mathfrak{m}\right] f$ shall denote the coefficient of
$\mathfrak{m}$ in $f$.

We equip $\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right]
$ with a topology, which is the product topology coming from viewing
$\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ as an
infinite direct product of copies of $\mathbf{k}$ (indexed by monomials).
Explicitly, this topology can be easily described in terms of limits of nets
(see Stijn Vermeeren's *Sequences and nets in
topology* for an
introduction to nets): A net $\left( f_{s}\right) _{s\in S}$ of power series
$f_{s}\in\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $
(where $S$ is a directed set) converges to a power series $f\in\mathbf{k}
\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ if and only if for
each monomial $\mathfrak{m}$, the net $\left( \left[ \mathfrak{m}\right]
f_{s}\right) _{s\in S}$ converges to $\left[ \mathfrak{m}\right] f$ in the
discrete topology on $\mathbf{k}$ (that is, $\left[ \mathfrak{m}\right]
f_{s}=\left[ \mathfrak{m}\right] f$ for all sufficiently high $s$, where the
specific meaning of "sufficiently high" depends on $\mathfrak{m}$).

This topology turns $\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3}
,\ldots\right] \right] $ into a topological $\mathbf{k}$-algebra. It is easy
to see that any $f\in\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3}
,\ldots\right] \right] $ satisfies
\begin{equation}
f=\lim\limits_{n\rightarrow\infty}f\left( x_{1},x_{2},\ldots,x_{n}
,0,0,0,\ldots\right)
\label{eq.darij1.1}
\tag{1}
\end{equation}
(since each monomial $\mathfrak{m}$ uses only finitely many indeterminates,
and thus remains unchanged when we replace all the higher indeterminates by
$0$). If $f$ is a symmetric function, then we conventionally write $f\left(
x_{1},x_{2},\ldots,x_{n}\right) $ for $f\left( x_{1},x_{2},\ldots
,x_{n},0,0,0,\ldots\right) $, so this equality becomes
\begin{align*}
f=\lim\limits_{n\rightarrow\infty}f\left( x_{1},x_{2},\ldots,x_{n}\right) .
\end{align*}

Infinite sums $\sum_{s\in S}f_{s}$ of summable families $\left( f_{s}\right)
_{s\in S}$ are easily defined using the above topology. Summability of a
family $\left( f_{s}\right) _{s\in S}$ means that the net $\left(
\sum_{s\in T}f_{s}\right) _{T\subseteq S\text{ finite}}$ converges, or,
equivalently, that for each monomial $\mathfrak{m}$, only finitely many $s\in
T$ satisfy $\left[ \mathfrak{m}\right] f_{s}\neq0$. In particular, if
$\left( f_{s}\right) _{s\in S}$ is a family of homogeneous power series, and
if it contains only finitely many degree-$n$ entries for each given
$n\in\mathbb{N}$, then it is summable. Thus, sums like $\sum_{\lambda
\in\operatorname*{Par}}s_{\lambda}$ and $\sum_{k\geq1}p_{k}$ are well-defined.

Now, it is easy to see that if $\left( f_{s}\right) _{s\in S}$ and $\left(
g_{t}\right) _{t\in T}$ are two summable families of power series satisfying
\begin{align*}
\sum_{s\in S}f_{s}\left( x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots\right)
=\sum_{t\in T}g_{t}\left( x_{1},x_{2},\ldots,x_{n},0,0,0,\ldots\right)
\end{align*}
for each $n\in\mathbb{N}$, then
\begin{equation}
\sum_{s\in S}f_{s}=\sum_{t\in T}g_{t}.
\label{eq.darij1.2}
\tag{2}
\end{equation}
Indeed, this follows by comparing the coefficient of each monomial
$\mathfrak{m}$ (noticing that $\mathfrak{m}$, once again, contains only the
indeterminates $x_{1},x_{2},\ldots,x_{n}$ for some $n\in\mathbb{N}$). Note
that I am not sure whether limits commute with summable sums in general, but
fortunately this is not needed to prove \eqref{eq.darij1.2}.

This almost suffices to derive your equality ($\star\star$) from your ($\star
$) (upon taking $\mathbf{k}=\mathbb{Q}\left[ p_{1},p_{2},p_{3},\ldots\right]
$), except that we need one more ingredient: The exponential. The definition
of the exponential is well-known: If $\mathbf{k}$ is a $\mathbb{Q}$-algebra,
and if $f\in\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right]
\right] $ is a power series with constant term $0$, then the exponential
$\exp f$ of $f$ is defined by
\begin{align*}
\exp f=\sum_{k\in\mathbb{N}}\dfrac{f^{k}}{k!},
\end{align*}
which is easily seen to be a sum of a summable family. Thus, we obtain a map
$\exp$
\begin{align*}
& \text{from }\left\{ \text{power series }f\in\mathbf{k}\left[ \left[
x_{1},x_{2},x_{3},\ldots\right] \right] \text{ with constant term
}0\right\} \\
& \text{to }\left\{ \text{power series }f\in\mathbf{k}\left[ \left[
x_{1},x_{2},x_{3},\ldots\right] \right] \text{ with constant term
}1\right\} .
\end{align*}
This map $\exp$ is furthermore continuous (because for each monomial
$\mathfrak{m}$, we can compute $\left[ \mathfrak{m}\right] \left( \exp
f\right) $ from knowing only the finitely many coefficients $\left[
\mathfrak{n}\right] f$ for the finitely many monomials $\mathfrak{n}$ that
divide $\mathfrak{m}$). As a consequence, it commutes with taking limits and
with taking sums of summable families. The substitution homomorphism
\begin{align*}
\mathbf{k}\left[ \left[
x_{1},x_{2},x_{3},\ldots\right] \right] &\to \mathbf{k}\left[ \left[
x_{1},x_{2},\ldots,x_{n}\right] \right] , \\
f &\mapsto f\left(x_1,x_2,\ldots,x_n,0,0,0,\ldots\right)
\end{align*}
is also continuous for each $n \in \mathbb{N}$, and commutes with $\exp$.
The rest is substitution.