The Lie derivative of any tensor field $K$ with respect to a vector field $\xi$ is by definition
$$L_\xi(K)=-\frac d{dt}|_{t=0}(\phi_t)_*K,$$
where $\phi_t$ is the local flow of $\xi$. Now, if $M$ has a Riemannian metric $g$ and $\xi$ is Killing with respect to $g$, each $\phi_t$ is a local isometry of $(M,g)$. From the uniqueness of the Levi-Civita connection, it follows that every isometry $\phi$ is affine, i.e.
$$\phi_*(\nabla_X Y)=\nabla_{\phi_*X}\phi_*Y.$$
From here you get immediately $\phi_*R=R$ for the Riemannian curvature, and since the Ricci tensor of $R$ is obtained by a trace:
$$Ric(X,Y)=trace(V\mapsto R_{V,X}Y),$$

one gets $\phi_*Ric=Ric$ for every isometry $\phi$. The first formula thus shows that $L_\xi Ric=0$ for every Killing vector field $\xi$.

**Edit**: Here is another, purely tensorial, proof of the same statement. Let $\xi$ be Killing, in the sense that $g(\nabla_X\xi,Y)+g(X,\nabla_Y\xi)=0$ for all vector fields $X,Y$. After taking the covariant derivative wrt some vector field $Z$, and doing some standard manipulations, one gets the usual Kostant formula:
$$\nabla^2_{X,Y}\xi=R_{\xi,X}Y,\qquad\forall X,Y\in C^\infty(TM).$$
This is just a rewriting of
$$L_\xi(\nabla_XY)=\nabla_{L_\xi X}Y+\nabla_X(L_\xi Y),$$
i.e. some sort of Leibniz formula.
Applying this formula several times eventually yields the corresponding Leibniz formula for $R$:
$$L_\xi(R_{X,Y}Z)=R_{L_\xi X,Y}Z+R_{X,L_\xi Y}Z+R_{X,Y}(L_\xi Z),$$
i.e. $L_\xi R=0$, and finally $L_\xi Ric=0$ after taking the trace.

Of course, this is just the infinitesimal version of the first proof...