Now I have found a proof. In the first jet space, it is sufficient to take $$\hat N^\mu=\tau^\mu+\sigma_a\,\frac{\partial}{\partial u_{a,\,\mu}},$$ and the proof proceeds as follows. We have
\begin{equation}
[\hat G+\tau^\nu_{,\,\nu},\,\hat N^\mu]=\hat G(\tau^\mu)+\hat G(\sigma_a)\,
\frac{\partial}{\partial u_{a,\,\mu}}+\sigma_a\left [\hat G,\,\frac{\partial}
{\partial u_{a,\,\mu}}\right]+\sigma_a\left [\tau^\nu_{,\,\nu},\,
\frac{\partial}{\partial u_{a,\,\mu}}\right].
\tag{1}
\end{equation}
But
\begin{equation}
\left [\tau^\nu_{,\,\nu},\,\frac{\partial}{\partial u_{a,\,\mu}}\right]=
-\frac{\partial \tau^\nu_{,\,\nu}}{\partial u_{a,\,\mu}}=-\frac{\partial
\tau^\mu}{\partial u_a},
\tag{2}
\end{equation}
because
\begin{equation}
\tau^\nu_{,\,\nu}=\frac{\partial \tau^\nu}{\partial x^\nu}+u_{b,\,\nu}\,
\frac{\partial \tau^\nu}{\partial u_b}.
\tag{3}
\end{equation}
On the other hand, as $\tau^\mu$ doesn't depend on field derivatives,
\begin{equation}
\hat G(\tau^\mu)=\tau^\nu\,\tau^\mu_{,\,\nu}+\sigma_a\,\frac{\partial
\tau^\mu}{\partial u_a}.
\tag{4}
\end{equation}
Further we have
\begin{equation}
\hat G(\sigma_a)=\tau^\nu\,\sigma_{a,\,\nu}+\sigma_b\,\frac{\partial
\sigma_a}{\partial u_b}+\sigma_{b,\,\nu}\,\frac{\partial \sigma_a}
{\partial u_{b,\,\nu}}.
\tag{5}
\end{equation}
But
$$\sigma_a=\xi(x,u)-\tau^\mu(x,u)\,u_{a,\,\mu},$$
and, therefore,
\begin{equation}
\frac{\partial \sigma_a}{\partial u_{b,\,\nu}}=-\delta_a^b\,\tau^\nu.
\tag{6}
\end{equation}
Substituting this into (5), we get
\begin{equation}
\hat G(\sigma_a)=\sigma_b\,\frac{\partial \sigma_a}{\partial u_b}.
\tag{7}
\end{equation}
It remains to calculate the commutator
\begin{equation}
\left [\hat G,\,\frac{\partial} {\partial u_{a,\,\mu}}\right]=
\left [\tau^\nu \,\frac{\partial }{\partial x^\nu}+\xi_b\,\frac{\partial }
{\partial u_b}+\eta_{b\nu}\,\frac{\partial }{\partial u_{b,\,\nu}},\,
\frac{\partial} {\partial u_{a,\,\mu}}\right]=-\frac{\partial \eta_{b\nu}}
{\partial u_{a,\,\mu}}\,\frac{\partial }{\partial u_{b,\,\nu}},
\tag{8}
\end{equation}
where we have used the fact that $\tau^\nu$ and $\xi_a$ do not depend on
field derivatives. Using
$$\eta_{b\nu}=\xi_{b,\,\nu}-\tau^\alpha_{,\,\nu}\,u_{b,\,\alpha},$$
along with
\begin{equation}
\frac{\partial \xi_{b,\,\nu}}{\partial u_{a,\,\mu}}=\delta^\mu_\nu\,
\frac{\partial \xi_b}{\partial u_a},\;\;\;\frac{\partial
\tau^\alpha_{,\,\nu}}{\partial u_{a,\,\mu}}=\delta^\mu_\nu\,
\frac{\partial \tau^\alpha}{\partial u_a},
\tag{9}
\end{equation}
we get
\begin{equation}
\frac{\partial \eta_{b\nu}}{\partial u_{a,\,\mu}}=\delta^\mu_\nu\,\left (
\frac{\partial \xi_b}{\partial u_a}-u_{b,\,\alpha}\,\frac{\partial
\tau^\alpha}{\partial u_a}\right )-\delta_a^b\,\tau^\mu_{,\,\nu}=
\delta^\mu_\nu\,\frac{\partial \sigma_b}{\partial u_a}-\delta_a^b\,
\tau^\mu_{,\,\nu}.
\tag{10}
\end{equation}
Therefore
\begin{equation}
\left [\hat G,\,\frac{\partial} {\partial u_{a,\,\mu}}\right]=
\tau^\mu_{,\,\nu}\,\frac{\partial }{\partial u_{a,\,\nu}}-
\frac{\partial \sigma_b}{\partial u_a}\,\frac{\partial }
{\partial u_{b,\,\mu}}.
\tag{11}
\end{equation}
Now (2), (4), (7) and (11), in combination with (1), imply the desired result:
\begin{equation}
[\hat G+\tau^\nu_{,\,\nu},\,\hat N^\mu]=\tau^\mu_{,\,\nu}\,\left (\tau^\nu+
\sigma_a\,\frac{\partial }{\partial u_{a,\,\nu}}\right )=\tau^\mu_{,\,\nu}\,
\hat N^\nu.
\end{equation}