The answer to your question is "yes". Let's start with a couple definitions of $X_1^{iso}$ that don't use a base scheme $S$:
It is the limit of the diagram made of the three arrows $i: X_0 \to X_1$, $m: X_1 \times_{X_0} X_1 \to X_1$, and $m \circ \sigma: X_1 \times_{X_0} X_1 \to X_1$, where $\sigma$ is the switching endomorphism of $X_1 \times_{X_0} X_1$. On the level of functors of points, this is the set of pairs $(f,g)$ such that $fg$, $gf$ and the identity coincide.
It is the fiber product of the endomorphism $(m, m \circ \sigma)$ on $X_1 \times_{X_0} X_1$ with the diagonal unit map $\Delta \circ i: X_0 \to X_1 \times_{X_0} X_1$. This mashes together two of the arrows from the above definition, and it is the analogue of Matt E's description from the Stackexchange answer.
There is a canonical monomorphism to $X_1$, taking $(f,g)$ to $f$. Also, we note that $X_1^{iso}$ is locally of finite presentation over $X_0$ by basic permanence properties (see e.g., Stacks 02KL with $X=S=X_1$ and $Y = X_0$). We will show that the monomorphism $X_1^{iso} \to X_1$ is formally étale by rephrasing Matt E's argument in a bit more generality.
Let $s: U \to T$ be a closed immersion of affine schemes over $X_0$ defined by a square zero ideal $I$. Let $\bar{f}$ be a $U$-point of $X_1^{iso}$, and let $f$ be a $T$-point of $X_1$, such that $f \circ s$ is equal to $\bar{f}$ followed by the monomorphism. By assumption, $\bar{f}$ has an inverse $\bar{g}$, and by smoothness of $X_1 \to X_0$, there exists some $T$-point $\tilde{g}$ of $X_1$ that restricts to the $U$-point $\bar{g}$.
We now translate by $\tilde{g}$ to bring us to an infinitesimal neighborhood of the origin. $\bar{g}$ and $\bar{f}$ are inverses as $U$-points, so $\tilde{g} f$ and $f \tilde{g}$ are $T$-points whose restriction to $U$ factors through identity. The square zero condition on $I$ gives an additive bijection between lifts of the identity and elements of $\operatorname{Hom}_{\mathcal{O}_U}(e_U^*\Omega_{X_1/X_0},I)$, where $e_U: U \to X_1$ is the identity section (SGAI Exp. 3 Proposition 5.1). Because the Hom space is an abelian group, we can easily form inverses by negating. Translating back yields $(\tilde{g} f)^{-1} \tilde{g} = \tilde{g} (f \tilde{g})^{-1}$ as the unique inverse of $f$.
To summarize, if $f$ is a $T$-point in $X_1$ whose restriction to $U$ factors through $X_1^{iso}$, then there is a unique $T$-point $(f,f^{-1})$ in $X_1^{iso}$ that maps to $f$. Thus, $X_1^{iso} \to X_1$ is formally étale, and the composition with either source or target $X_1 \to X_0$ is formally smooth and locally of finite presentation, hence smooth.
This argument works for algebraic spaces without change (although you may need to find new references).