I claim that the answer to your question is "yes" if $X_1$ is separated and umramified over our base scheme $S$. I'm not saying that that this (strong?) assumption is essential. It is just precisely what one needs in my easy proof:

First we have to define $X_1^{iso}$. Consider the fiber product $X_1 \times_{X_0 \times_S X_0} X_1$ with respect to $(t,s) : X_1 \to X_0 \times_S X_0$ and $X_0 \times_S X_0 \leftarrow X_1 : (s,t)$. Intuitively (and this holds verbatim in the functorial picture), it consists of morphisms $(f,g)$ such that we can compose $f \circ g$ *and* $g \circ f$. Define a morphism $\alpha : X_1 \times_{X_0 \times X_0} X_1 \to X_1 \times X_1 \times X_1 \times X_1$ by $\alpha(f,g)=(i(s(f)),i(s(g)),m(f,g),m(g,f))$, where $i : X_0 \to X_1$ is the unit and $m : X_1 \times_{t,X_0,s} X_1 \to X_1$ is the composition. Besides, we have the diagonal $\Delta_{X_1 \times X_1}$ mapping $(a,b) \mapsto (a,b,a,b)$. Then we define $X_1^{iso}$ by the cartesian square

$$\begin{matrix} X_1^{iso} & \rightarrow & X_1 \times X_1 \\\\ \downarrow j & & ~~ \downarrow\Delta_{X_1 \times X_1} \\\\ X_1 \times_{X_0 \times X_0} X_1 & \xrightarrow{\alpha} & X_1 \times X_1 \times X_1 \times X_1 \end{matrix}$$

Then $X_1^{iso}$ consists of pairs of morphisms $(f,g)$ such that $f \circ g$ and $g \circ f$ are well-defined and equal the identity. The map $X_1^{iso} \to X_1$ is defined as the composition
$$X_1^{iso} \xrightarrow{j} X_1 \times_{X_0 \times X_0} X_1 \xrightarrow{\mathrm{pr_1}} X_1.$$
In general, this is a locally closed immersion. Now $X_1$ is separated, hence the same is true for $X_1 \times X_1$ and its diagonal is a closed immersion. Hence, $j$ is a closed immersion. Now, if I am not mistaken, in general $\Delta_X$ is formally smooth iff $X$ is formally unramified, and $\Delta_X$ is of finite presentation when $X$ is of finite presentation. Hence, by our assumption, $j$ is even smooth. And $\mathrm{pr}_1$ is some base change of the morphism $(s,t) : X_1 \to X_0 \times X_0$, which is smooth since it is the smooth diagonal $\Delta_{X_1}$ followed by the smooth morphism $s \times t : X_1 \times X_1 \to X_0 \times X_0$.

This shows that $X_1^{iso} \to X_1$ is smooth. In particular, the compositions $s,t : X_1^{iso} \to X_1 \rightrightarrows X_0$ are smooth.