The answer to your question is "yes". Let's start with a couple definitions We need the following properties of $X_1^{iso}$ that don't use a base scheme $S$:
ItThe map $X_1^{iso} \to X_1$ given by $(f,g) \mapsto f$ on scheme-valued points is the limita monomorphism. This is straightforward from uniqueness of the diagram madeinverses.
$s,t: X_1 \to X_0$ are formally smooth. We use this for existence of the three arrowsa lift of an infinitesimal deformation.
$s,t: X_1 \to X_0$ are locally of finite presentation. Using basic permanence properties $i: X_0 \to X_1$(see e.g., $m: X_1 \times_{X_0} X_1 \to X_1$,Stacks 02KL with $X=S=X_1$ and $m \circ \sigma: X_1 \times_{X_0} X_1 \to X_1$$Y = X_0$), wherewe find that $\sigma$$i: X_0 \to X_1$ is the switching endomorphismalso locally of $X_1 \times_{X_0} X_1$finite presentation. On Combining with the level of functors of pointsfiber product construction, this is the set of pairs $(f,g)$ such thattwo maps from $fg$,$X_1^{iso}$ to $gf$ and the identity coincide$X_0$ are also lfp.
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 $\Delta \circ i: X_0 \to X_1 \times_{X_0} X_1$Infinitesimal formal categories are formal groupoids. This mashes together two of the arrows from the above definition, and it is the analogue of Matt E's description frombasically the Stackexchange answerstatement that tangent spaces have abelian group structure.
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 $Y = X_0$). WeWe 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$$j: U \to T$ be a closed immersion of affine schemes over $X_0$ defined by a square zero ideal $I$. Let $\bar{f}$ be a $U$-point of $X_1^{iso}$, and let $f$ be a $T$-point of $X_1$, such that $f \circ s$$f \circ j$ is equal to $\bar{f}$ followed by the monomorphism. By assumption, $\bar{f}$ has an inverse $\bar{g}$, and by formal smoothness of both $X_1 \to X_0$$s,t: 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).