The linear operator $P$ depends on $t$, and we have $(1)$ as long as $P(t) = 0$. This is why we need to know more about $P$ to conclude that $t$ is not a root of the Bernstein-Sato polynomial whenever $t > -lct(f)$.

Edit notice: The answer is completely rewritten due to user2520938's comment. My original answer was that the linear operator $P$ depends on $t$, and we have $(1)$ as long as $P(t) = 0$. But as user2520938 pointed out, we always have $P(t) \ne 0$, since otherwise it contradicts the minimality of the Bernstein-Sato polynomial. So we need an example with $P(t) \ne 0$.

Here are some examples with $t = \frac{1}{k} - 1$ where $k$ is a positive integer. In the one-variable situation, consider $f(z) = z^k$ and define $$P(\phi) := 2\phi + z\frac{\partial\phi}{\partial z}.$$ We have $$ \begin{split} \int\int |f(z)^2|^{1/k}(P \bar{P}(\phi(z))) dxdy = \int \int \frac{\partial^2}{\partial z\partial \bar{z}}\left(z^2\bar{z}^2\phi \right) dxdy \\ \int \int \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)\left( (x^2 + y^2)^2\phi\right) dxdy \end{split} $$ with $x + iy = z$. From this expression, we see that if $\phi$ has compact support, then the integral vanishes.

Obviously $P$ is not a differential operator which characterizes the Bernstein-Sato polynomial of $f$, and this example is constructed without assuming that $P$ is a priori related to $f$.

The linear operator $P$ depends on $t$, and we have $(1)$ as long as $P(t) = 0$. This is why we need to know more about $P$ to exclude that $t$ is not a root of the Bernstein-Sato polynomial whenever $t > lct(f)$.

The linear operator $P$ depends on $t$, and we have $(1)$ as long as $P(t) = 0$. This is why we need to know more about $P$ to conclude that $t$ is not a root of the Bernstein-Sato polynomial whenever $t > -lct(f)$.