I am wondering if there is a simple example of a manifold such that, given a value for the scalar curvature $R$, I can find a manifold such that the Ricci tensor has all zero components except for one component which takes the value $R$.
I feel like this can be achieved using a warped product of two metrics to separate out one coordinate and then just solve the differential equations so that the first coefficient vanishes, but obviously the coefficient of the non-zero component needs to be $R$.
Your intuition is correct that warped product metrics will do the trick.
Maybe try to redo the computations yourself, but if I'm not mistaken the Ricci curvature of $g=dt^2+f(t)^2dx^2$ (where $x\in\mathbb{R}^n$ and $dx^2$ denotes the euclidean metric on $\mathbb{R}^n$) is given by : $$\text{Ric}_g=-(n-1)\frac{f''(t)}{f(t)}dt^2-\left(f''(t)f(t)+(n-2)f'(t)^2\right)dx^2.$$ So any positive non constant solution to $f''f+(n-2)(f')^2=0$ will give you a solution, these are given by : $f(t)=C_2(C_1+(n-1)t)^{\tfrac{1}{n-1}}$. However note that the scalar curvature will not be constant in that example and that the singularity which happens when $f$ vanishes is genuine (the scalar curvature will blow up).
Maybe we can get something better (smooth for instance) using $\mathbb{R}\times \mathbb{S}^n$ or $\mathbb{R}\times \mathbb{H}^n$, this way the ODE becomes $f''f+(n-2)(f')^2=\pm 1$. The trick which solved the ODE in the first case doesn't work here though and a more careful analysis is required.
I don't know if one can find less symmetric examples, maybe studying the vector field given by the eigenvector of the Ricci tensor associated to the nonzero eigenvalue can give something.
