It seems false to me without assuming the space to be $\sigma$-finite: take $\Omega:=[0,1]^2$, with $\mu:=\mathcal{H}^1$ (1-dimensional Hausdorff measure) and the $\sigma$-algebra $\mathcal A$ generated by horizontal and vertical slices ($\{s\}\times[0,1]$ and $[0,1]\times\{t\}$).

Now with little work you can show that all elements of $\mathcal A$, up to adding and removing negligible sets, are of the form
$$\bigcup\Big(\{s_i\}\times[0,1]\Big)\cup\bigcup\Big([0,1]\times\{t_j\}\Big),$$
where both unions are (at most) countable. Hence, $L^2(\mu)$ splits as a direct sum $V\oplus W$, where $V$ consists of functions of the form $f=\sum_i a_i 1_{\{s_i\}\times[0,1]}$ and $W$ of similar "vertical" functions.

Now declare $T$ to act by multiplication by $0$ on $V$ and multiplication by $1$ on $W$. It's easy to see that there is no consistent choice of $v$.

If you don't want atoms in the counterexample, take instead the $\sigma$-algebra generated by sets of the form $\{s\}\times E'$ and $E\times\{t\}$, where $s,t$ range in $[0,1]$ and $E,E'$ vary among Borel subsets of $[0,1]$. In this case, measurable sets have the form
$$\bigcup(\{s_i\}\times E_i')\cup\bigcup(E_j\times\{t_j\})\cup (E\times E')\cup N,$$
where $E_j,E$ are Borel subsets of $[0,1]\setminus\bigcup\{s_i\}$, $E_i',E'$ are Borel subsets of $[0,1]\setminus\bigcup\{t_j\}$, and finally $N$ is any subset of $\Big(\bigcup\{s_i\}\Big)\times\Big(\bigcup\{t_j\}\Big)$.
Once you have this, you can argue as before.