polynomial time reducibility - $L_{2} \notin \textbf{P}$ and $L_{1} \leq_{p} L_{2} \implies L_{1} \notin \textbf{P}$
Let $L_1=\emptyset$ and $L_2=\{\langle M\rangle\mid M\text{ halts on empty string}\}$. Let $s$ be a string that does not belong to $L_2$, for example, the empty string, which is a string that does not even represent a Turing machine. The reduction $f$ is $\forall x, f(x)=s$.

Now you can see $L_1\in\mathbf{P}$ and $L_2\notin\mathbf{P}$ (in fact, $L_2$ is undecidable), and $f$ is a polynomial-time reduction (it just ignores the input and outputs $s$ directly). Also, $x\in L_1\Leftrightarrow f(x)\in L_2$ (note $x$ always $\notin$ $L_1$ and $f(x)=s$ always $\notin$ $L_2$).

This is a counter-example.