Asssume that the following integral exists. Does there exists a non-trivial bounded measurable function $f(x)$ such that for all $y>0$, $$\int_0^\infty f(x) e^{-xy} dx=0?$$
If the the function $f(x)$ satisfies conditions(like being of exponential order)of Laplace transform, then $f(x)=0$ by the uniqueness of Laplace transforms.
Saying that $\int_0^{\infty}e^{-xy}f(x)dx=0$ for all $y>0$ postulates that $\int_0^{\infty}e^{-xy}|f(x)|dx<\infty$ for all $y>0$. With the notation $f=f^+-f^-$ (with $f^+=\max (f,0)$ and $f^-=(-f)^+$) we have $$L_+(y)=\int_0^{\infty}e^{-xy}f^+(x)dx=\int_0^{\infty}e^{-xy}f^-(x)dx=L_-(y)$$ One can prove without great effort that $L_{\pm}$ are extendable to two unique analytic functions on $H=\{z; \Re z>0\}$. Since they coincide on the real line, from the principle of isolated zeros they coincide on $H$. In particular $L_+(y_0+it)=L_-(y_0+it)$ for all real $t$, which means that the two Fourier transforms of the bounded positive measures $e^{-xy_0}f^{\pm}(x)dx$ coincide and from the Paul Levy theorem the measures themselves coincide. Therefore $f^+=f^-$ almost everywhere, meaning $f=0$ almost everywhere.
Since there is no condition on $f(x)$ for $x \lt 0$ in your question, for $f(x) = g(x) \theta(-x)$, where $\theta$ is Heaviside function and $g(x)$ is to be constructed to satisfy boundedness, non-trviality and measurablility, your integral
$$\int\limits_{0}^{\infty} f(x) e^{-xy} dx = \int\limits_{0}^{\infty} g(x) \theta(-x) e^{-xy} dx = 0$$ for $y \gt 0$.
Clearly $f(x)$ is not identically $0$ over the real axis.
However, if you were to edit your question to add that $f(x) = 0$ for $x \lt 0$ as an additional condition, comment from Mittens is valid. Laplace Transform is analytic in the region of convergence. If we can prove that no analytic function exists such that it is identically zero on the positive x-axis but non-zero elsewhere in the complex plane, it would rigorously prove the invertibility of LT would ensure that $f(x) = 0$.
