This answer gives some partial progress towards your conjecture. Since the Coxeter polynomial is also a derived invariant, it should have the same applications.

Theorem: Suppose $D$ is a Dyck path of length $n$. The Coxeter polynomial (characteristic polynomial of $\phi_{D}$) is equal to $\frac{x^{n+1}-1}{x-1}$ if and only if $D$ is bouncing.

Before we get to the proof, let's set up some notation. We denote by $(u_1,v_1), (u_2,v_2), \dots, (u_k,v_k)$ the coordinates of the valleys of $D$. These are the coordinates of the entries in the Cartan matrix of $D$ which are (1) above the diagonal (2) equal to zero (3) the entries directly below and directly to the left are equal to 1. For example, the valleys of $[2,5,4,3,3,2,1]$ are $(1,3),(4,7)$.

Next, we define the matrix $X_D$ similarly to the Cartan matrix except we put ones in coordinates $(i,i+c_i-1)$ and zeros everywhere else. So $X_D$ essentially consists of just the "righmost" 1's in the Cartan matrix. The matrix $Y_D$ is defined as the matrix with $-1$'s in positions $(u,u+1),(u,u+2),\dots, (u,v)$ as $(u,v)$ ranges through all the valleys, and zeros everywhere else. Finally let $A_n$ be the matrix with 1's in entries $(i,i+1)$ for $i=1,\dots,n-1$. One can check the following explicit form for the Coxeter matrix of a Dyck path: $$\phi_D=A_n+Y_D-X_D^{\top}.$$ We will also need a lemma

Lemma: The characteristic polynomial of $\phi_D$ is equal to $x^n+x^{n-1}+(1+\alpha) x^{n-2}+O(x^{n-3})$. Where $\alpha$ is the number of valleys $(u,v)$ of $D$ with $v>u+2$.

Proof of Lemma: We get $x^{n}+x^{n-1}$ by looking at the product of elements in the diagonal of $xI-\phi_D$. The coefficient of $x^{n-2}$ is precisely $-\sum_{i<j}\phi_D(i,j)\phi_D(j,i)$. By analyzing the explicit form for $\phi_D$ above, we have $\phi_D(i,j)\phi_D(j,i)=-1$ when either $(i,j)=(n-1,n)$ or when $(i,j)=(u-1,v)$ and $(u,v)$ is a valley with $u+2<v$, and $\phi_D(i,j)\phi_D(j,i)=0$ otherwise.

Proof of Theorem: The lemma tells us that the characteristic polynomial of a path which is not bouncing cannot be equal to $\frac{x^{n+1}-1}{x-1}$. To show that every bouncing path has this as the characteristic polynomial you can use induction on the length of the Dyck path. If $D'$ is the Dyck path corresponding to the sequence $[c_2,c_3,\dots,c_n]$ then $D'$ is also bouncing, and $\phi_{D'}$ is the $(1,1)$ cofactor of $\phi_D$. By expanding the determinant along the first row one can establish the recurrence $$\det(xI_n-\phi_D)=1+x\det(xI_{n-1}-\phi_{D'})$$ (One needs to show that there is a unique permutation not fixing 1, that has nonzero contribution in the determinant expansion.) This gives us our result when combined together with the inductive claim that the Coxeter polynomial of $D'$ is $\frac{x^n-1}{x-1}$.

This answer gives some partial progress towards your conjecture. Since the Coxeter polynomial is also a derived invariant, it should have the same applications.

Theorem: Suppose $D$ is a Dyck path of length $n$. The Coxeter polynomial (characteristic polynomial of $\phi_{D}$) is equal to $\frac{x^{n+1}-1}{x-1}$ if and only if $D$ is bouncing.

Before we get to the proof, let's set up some notation. We denote by $(u_1,v_1), (u_2,v_2), \dots, (u_k,v_k)$ the coordinates of the valleys of $D$. These are the coordinates of the entries in the Cartan matrix of $D$ which are (1) above the diagonal (2) equal to zero (3) the entries directly below and directly to the left are equal to 1. For example, the valleys of $[2,5,4,3,3,2,1]$ are $(1,3),(4,7)$.

Next, we define the matrix $X_D$ similarly to the Cartan matrix except we put ones in coordinates $(i,i+c_i-1)$ and zeros everywhere else. So $X_D$ essentially consists of just the "righmost" 1's in the Cartan matrix. The matrix $Y_D$ is defined as the matrix with $-1$'s in positions $(u,u+1),(u,u+2),\dots, (u,v)$ as $(u,v)$ ranges through all the valleys, and zeros everywhere else. Finally let $A_n$ be the matrix with 1's in entries $(i,i+1)$ for $i=1,\dots,n-1$. One can check the following explicit form for the Coxeter matrix of a Dyck path: $$\phi_D=A_n+Y_D-X_D^{\top}.$$ We will also need a lemma

Lemma: The characteristic polynomial of $\phi_D$ is equal to $x^n+x^{n-1}+(1-\alpha) x^{n-2}+O(x^{n-3})$. Where $\alpha$ is the number of valleys $(u,v)$ of $D$ with $v>u+2$.

Proof of Lemma: We get $x^{n}+x^{n-1}$ by looking at the product of elements in the diagonal of $xI-\phi_D$. The coefficient of $x^{n-2}$ is precisely $-\sum_{i<j}\phi_D(i,j)\phi_D(j,i)$. By analyzing the explicit form for $\phi_D$ above, we have $\phi_D(i,j)\phi_D(j,i)=-1$ when $(i,j)=(n-1,n)$ and we have $\phi_D(i,j)\phi_D(j,i)=1$ when $(i,j)=(u-1,v)$ and $(u,v)$ is a valley with $u+2<v$, and $\phi_D(i,j)\phi_D(j,i)=0$ otherwise.

Proof of Theorem: The lemma tells us that the characteristic polynomial of a path which is not bouncing cannot be equal to $\frac{x^{n+1}-1}{x-1}$. To show that every bouncing path has this as the characteristic polynomial you can use induction on the length of the Dyck path. If $D'$ is the Dyck path corresponding to the sequence $[c_2,c_3,\dots,c_n]$ then $D'$ is also bouncing, and $\phi_{D'}$ is the $(1,1)$ cofactor of $\phi_D$. By expanding the determinant along the first row one can establish the recurrence $$\det(xI_n-\phi_D)=1+x\det(xI_{n-1}-\phi_{D'})$$ (One needs to show that there is a unique permutation not fixing 1, that has nonzero contribution in the determinant expansion.) This gives us our result when combined together with the inductive claim that the Coxeter polynomial of $D'$ is $\frac{x^n-1}{x-1}$.