I was wondering if anyone had a simple example of a manifold which only has Ricci curvature in one direction ie. such that the Ricci tensor only has one non-zero component.
Intuitively, I would expect such a thing to be possible somehow just from the definition of Ric but could not think of an example.
Yes, this is possible. More precisely: In dimension $\geq 3$ there are metrics for which the Ricci curvature has rank $1$, that is, which can be written (locally, around any point) as $\phi \alpha \otimes \alpha$ for some smooth function $\phi$ and $1$-form $\alpha$. (In dimension $2$ the Ricci tensor is a (smooth) multiple of the metric, so this behavior is possible only in dimension $\geq 3$.)
Here's one way to construct an example. The warped product of the flat metric $\bar g := dx_1^2 + \cdots + dx_{n - 1}^2$ on $\Bbb R^{n - 1}$ with the flat metric $dy^2$ on $\Bbb R$ via $f(y)$ is $$g := \bar g \times_{f(y)} dy^2 = f(y) \bar g + dy^2 .$$
Computing gives that the Ricci curvature of $g$ is $$\operatorname{Ric} = - \frac{2 f''(y) f(y) + (n - 1) f'(y)^2}{4 f(y)} \bar g - \frac{(n - 1)(2 f''(y) f(y) - f'(y)^2)}{4 f(y)^2} dy^2$$
Demanding that the coefficient of $\bar g$ vanish defines a second-order differential equation in $f$ with general solution $$f(y) = \left[\frac{n - 1}{2} (A y + B)\right]^{2 / (n - 1)},$$ and substituting in $g$ gives (now restricting the metric to where $A y + B > 0$) that $$\operatorname{Ric} = \frac{(n - 2) A^2}{(n - 1) (A y + B)} dy^2 .$$ So, if we take $A \neq 0$, then for $n > 2$, $\operatorname{Ric}$ has rank $1$.