Using experiments with QPA, it seems a counterexample has been found. It is quite complicated and surprisingly no simple counterexample seems to work. Maybe someone is interested to check it (I hope there is no mistake). There might be a simplificiation or even a simpler counterexample.
Let $A=KQ/I$ be the finite dimensional quiver algebra over a field $K$ where
\begin{tikzcd}
2 \arrow[r, "z", shift left=0.75ex] & 1 \arrow[loop, distance=3em, out=35, in=-35, "x"] \arrow[l, "y", shift left=0.75ex]
\end{tikzcd}
(how to use tikz here?)
$Q$ is the quiver with 2 vertices 1 and 2 and there is a loop x from 1 to 1, an arrow y from 1 to 2 and an arrow z from 2 to 1.
The relations are $I=\langle xy, yz, zx, x^3 \rangle$. The algebra is representation-finite and special biserial with 13 indecomposable modules.
Let $S_i$ denote the simple $A$-modules.
$A$ has vector space dimension 7 over the field $K$. Let $e_i$ denote the primitive idempotents of the quiver algebra $A$ corresponding to the vertices $i$ for $i=1,2$.
The indecomposable projective $A$-module $P_1=e_1 A= \langle e_1, x ,y , x^2 \rangle$ has dimension vector $[3,1]$ and the indecomposable projective $A$-modules $P_2=e_2 A= \langle e_2, z, zy \rangle $ has dimension vector $[1,2]$. We remark that $P_2$ is also injective.
For finite dimensional algebras we have $D Tr=\tau$, the Auslander-Reiten translate and since $D$ is a duality, the subcategory $Tr (\Omega^2(\mod-A))$ will be extension closed if and only if $\tau ( \Omega^2(\mod-A))$ is extension closed. We will thus look at the subcategory $\tau ( \Omega^2(\mod-A))$ in the following.
We use the notation $\tau_i:=\tau(\Omega^{i-1})$ in the following for the higher Auslander-Reiten translates.
Let $I=(x+z)A$ denote the right ideal generated by $x+z$, which has vector basis basis $\langle x+z, x^2, zy\rangle$.
$M_1:=A/I$ is an indecomposable $A$-module with dimension vector $[2,2]$.
Then it is easy to see that $\tau_3(M_1) \cong M_1$ and thus $M_1 \in \tau ( \Omega^2(\mod-A))$.
Now let $M_2:=P_2/S_2= e_2 A/zy A$, which is an indecomposable $A$-module with dimension vector $[1,1]$.
Again it is easy to see that $\tau_3(M_2) \cong M_2$ and thus $M_2 \in \tau ( \Omega^2(\mod-A))$.
Now $\dim(Ext_A^1(M_2,M_1))=1$ and there is (up to isomorphism) a unique short exact sequence that is not split:
$$0 \rightarrow M_1 \rightarrow W \rightarrow M_2 \rightarrow 0.$$
If we show that $W$ is not in $\tau ( \Omega^2(\mod-A))$ then we have shown that the subcategory $\tau ( \Omega^2(\mod-A))$ is not extension closed.
Now $W \cong P_2 \oplus U$, where $U=A/((x+y+z)A)$.
Here the ideal $(x+y+z)A$ has vector space basis $\langle x+y+z,y,x^2,zy \rangle$ and thus $U$ has dimension vector $[2,1]$.
Now let $f_1: P_2 \rightarrow M_2$ be the projective cover of $M_2$ and let
$f_2: U \rightarrow M_2$ be the canonical non-zero map $A/(xA+yA+zA) \rightarrow e_2 A/zyA=M_2$, which is given by the canonical surjections $g: A \rightarrow e_2 A \rightarrow e_2 A/zyA$ and noting that here $(xA+yA+zA)$ is in the kernel of $g$ which induces the map $f_2$.
Let $f:=f_1 \oplus f_2$, then it is easy to see that the kernel of $f$ is isomorphic to $M_1$ and since $W$ is not isomorphic to $M_1 \oplus M_2$ the short exact sequence is not split.
Now we show that the direct summand $U$ of $W$ is not in $\tau ( \Omega^2(\mod-A))$ or equivalently that $K:=\tau^{-1}(U)$ is not in $\Omega^2(\mod-A)$.
Now a module $T$ is a direct summand of a module of the form $X \oplus P$ for $X$ an $n$-th syzygy module and a projective module $P$ if and only if $T$ is a direct summand of a module of the form $P' \oplus \Omega^n(\Omega^{-n}(T))$.
Now for $T=K$, the module $\Omega^2(\Omega^{-2}(T))$ is isomorphic to $S_2 \oplus V$, where $V$ is an indecomposable $A$-module with dimension vector $[2,0]$.
Thus $K$ is not a direct summand of $P' \oplus \Omega^2(\Omega^{-2}(K))$ for any projective module $P'$ and thus $K$ is not a 2-th syzygy module.
Thus the subcategory $\tau ( \Omega^2(\mod-A))$ is not extension-closed.
