No. The problem is that $\tau-1$ is not a "stopping time" for the martingale.
Example. Martingale $(X_n), n=0,1,2$ is the standard random walk on $\mathbb Z$. Our sample space is $[0,1)$ with Lebesgue measure.
\begin{align}
X_0(\omega) &= 0,\qquad \omega \in [0,1).
\\
X_1(\omega) &= \begin{cases}
1,\quad & \omega \in [0,1/2),
\\
-1,\quad & \omega \in [1/2,1).
\end{cases}
\\
X_2(\omega) &= \begin{cases}
2,\quad & \omega \in [0,1/4),\\
0,\quad & \omega \in [1/4,1/2),\\
0,\quad & \omega \in [1/2,3/4),\\
-2,\quad & \omega \in [3/4,1).
\end{cases}
\end{align}
The filtration $\mathcal F_0, \mathcal F_1, \mathcal F_2$ for this martingale:
$\mathcal F_0$ is the sigma-algebra generated by $X_0$, so it is the trivial sigma-algebra with atom $[0,1)$.
$\mathcal F_1$ is the sigma-algebra generated by $X_0, X_1$, so it is the sigma-algebra with atoms $[0,1/2), [1/2,1)$.
$\mathcal F_2$ is the sigma-algebra generated by $X_0, X_1, X_2$, so it is the sigma-algebra with atoms $[0,1/4),[1/4,1/2),[1/2,3/4),[3/4,1)$.
Define stopping time $\tau = 2 \wedge \min\{n : X_n < 0\}.$ Then
\begin{align}
\tau(\omega) &= \begin{cases}
2,\quad & \omega \in [0,1/4),\\
2,\quad & \omega \in [1/4,1/2),\\
1,\quad & \omega \in [1/2,3/4),\\
1,\quad & \omega \in [3/4,1).
\end{cases}
\\
\sigma(\omega) &= \begin{cases}
1,\quad & \omega \in [0,1/4),\\
1,\quad & \omega \in [1/4,1/2),\\
0,\quad & \omega \in [1/2,3/4),\\
0,\quad & \omega \in [3/4,1).
\end{cases}
\end{align}
Now $\tau$ is a stopping time, so of course $\mathbb E[X_\tau] = 0$. But compute $\mathbb E[X_\sigma] = 1/2 \ne 0$, which shows that $\sigma$ is not a stopping time. Computations:
\begin{align}X_\tau(\omega) &= \begin{cases}
2,\quad & \omega \in [0,1/4),\\
0,\quad & \omega \in [1/4,1/2),\\
-1,\quad & \omega \in [1/2,3/4),\\
-1,\quad & \omega \in [3/4,1).
\end{cases}
\\
X_\sigma(\omega) &= \begin{cases}
1,\quad & \omega \in [0,1/4),\\
1,\quad & \omega \in [1/4,1/2),\\
0,\quad & \omega \in [1/2,3/4),\\
0,\quad & \omega \in [3/4,1).
\end{cases}
\end{align}
Calculation that shows $\sigma$ is not a stopping time:
$$
\{\sigma = 0\} = [1/2,1) \notin \mathcal F_0.
$$