Here is a big class of negative examples to your question. Let $A$ be a non-Gorenstein local Artinian ring and $M=w_A$ the canonical module of $A$.

Then $ann(M)=0$, as can be seen because $Hom(M,M)\cong A$. Suppose $M$ contains $s$ with annihilator $=(0)$. Then $As\cong A$ sits inside $M$. But since the length of $M$ is equal to the length of $A$, $M=As\cong A$, contradicting our choice of a non-Gorenstein $A$.

Simplest concrete example is $A=k[[x,y]]/(x,y)^2$.

Here is a big class of negative examples to your question. Let $A$ be a non-Gorenstein Artinian ring and $M=w_A$ the canonical module of $A$.

Then $ann(M)=0$, as can be seen because $Hom(M,M)\cong A$. Suppose $M$ contains $s$ with annihilator $=(0)$. Then $As\cong A$ sits inside $M$. But since the length of $M$ is equal to the length of $A$, $M=As\cong A$, contradicting our choice of a non-Gorenstein $A$.

Simplest concrete example is $A=k[[x,y]]/(x,y)^2$.

In fact, in the Artinian case one has

Proposition: For an Artinian ring $A$, the following are equivalent:

1. $A$ is Gorenstein.
2. Any finitely generated faithful module $M$ over $A$ contains an element with $(0)$ annihilator.

Proof: (2) implies (1) is above. Assume (1) and let $M$ be a faithful module. Let $s_1,...,s_n$ be a set of generators of $M$. Then we have an injection $A\to M^n$ taking $1$ to $(s_1,...,s_n)$. Since $A$ is Gorenstein, this injection splits. As Krull-Schmidt holds over $A$, $M$ contains $A$ as a free summand, which implies what we need.