Given a vector bundle $E\to M$ equipped with an anchor map $\rho:E\to TM$, choose a connection $\nabla$ on $E$ and consider the ODE
\begin{align*}
\dot\gamma(t) &= \rho(v(t))\\
(\gamma^*\nabla)_t v(t) &= 0
\end{align*}
for $\gamma:I\to M$ a path in $M$ and $v\in\Gamma(\gamma^*E)$ a section of the pulled back vector bundle. Standard results imply that for $(x_0,v_0)\in E$, there is an open interval $I(x,v)$ containing $0$ such that there is a unique solution $(x,v)$ of this ODE on $I$ with $(x(0),v(0)) = (x_0,v_0)$ and that $I(x,v)$ is maximal with respect to this property.
Let $U\subset E$ be the (open) subset of elements such that $1\in I(x,v)$, and note that it contains the zero section since constant paths with $v\equiv 0$ define a global solution of the ODE; then we have an obvious bijection between $U$ and germs of solutions $(x(t),v(t))$ of the ODE on $[0,1]$. In particular, the involution $(x(t),v(t))\mapsto (x(1-t),-v(1-t))$ on the latter space gives rise to an involution $\Phi:U\cong U$ fixing the zero section, and standard results show that it is smooth (it is essentially given by sending an element $(x,v)$ to the time $1$ value of the ``flow'' generated by it).
We obtain a reflexive, involutive graph by setting
- $G = U$
- $s = p:G\to M$ the bundle projection
- $t = p\circ \Phi:G\to M$
- $e:M\to G$ the inclusion of the zero section
- Lastly, the involution is given by $i = \Phi:U\cong U$.
It is straightforward, if tedious, to verify that applying your construction to this data gives back $E\xrightarrow{\pm \rho}M$ (the sign depends on some conventions and can be absorbed into the definition of the ODE).
This construction is essentially given by integrating the $L_\infty$-algebroid generated by $E\xrightarrow{\rho}M$ to an $L_\infty$-groupoid $\Delta^{op}\to \operatorname{Man}$ and restricting to $\Delta_{\le 1}^{op}$.
