If I define a general metric tensor for dimensions $d\geq 2$, does this imply there exists a (pseudo-)Riemannian manifold which admits this metric? I am specifically interested in Lorentzian manifolds, but also in the general case.
EDIT: I am specifically considering the metric tensor $$g_{\mu\nu}=\begin{bmatrix}-1 & 0 \\ 0 & \sinh t\end{bmatrix}$$
using the coordinates $t\in[0,\infty)$ and $\theta\in[0,2\pi)$.
You are mixing up the notions of manifold and Riemannian structure. By definition, a metric can only be defined on a manifold, and this metric defines the Riemannian structure. You have things backwards.
In your case, you already have your manifold, in this case a 2-dimensional half-cylinder with its border (note that it is thus not a manifold, but a manifold with border), which is a $\mathcal{C}^\infty$ manifold with its standard topology/atlas.
Then you get to define your "metric" (in your case it is not actually pseudo-Riemannian, as it will be degenerate on the border of your manifold). This gives your manifold the additional metric structure.