In general relativity, are two pseudo-Riemannian manifolds physically equivalent if they are isometric, or just diffeomorphic?

Indeed, "diffeomorphism invariance" of GR in physics in this context means in proper mathematical parlance that isometric (pseudo-)Riemannian manifolds are physically equivalent. In my view, this confusion between "diffeomorphism" and "isometry" is probably due to physicists usually looking at a manifold in coordinates, and a change in coordinates can be understood as an autodiffeomorphism from the manifold to itself, where the source carries one choice of coordinate charts and the target another.

Under such a change, the metric and all fields transform naturally via pushforward, and we essentially define the diffeomorphism to be an isometry so that the fields on the target with the new coordinates are equivalent to the fields on the source. So the slogan is "change of coordinates" = "diffeomorphism invariance", but the nature of the coordinate change means that the diffeomorphism is always additionally promoted to an isomorphism in whatever category of manifolds we are currently moving in.

This "diffeomorphism invariance" is emphatically not a special property of GR: Every proper physical theory does not care for the coordinates we choose. $\phi^4$-theory and Yang-Mills theory are precisely as diffeomorphism invariant in this sense as GR, just that there the diffeomorphism pushes forward not the metric, but a scalar field and a gauge connection, respectively.

Unfortunately, the phrase "diffeomorphism invariance" is occasionally also used in a different context, namely for what is more properly the local Lorentz invariance of GR (at least in the spin connection formalism). To wit, a diffeomorphism induces a map on all tensors via its Jacobian (or, more mathematically, the (co)tangent map) and GR also exhibits a invariance under the transformations from this (co)tangent map alone, i.e. it is a $\mathrm{GL}(n-1,1)$ (or $\mathrm{SO}(n-1,1)$) gauge theory, albeit an unusual one whose principal bundle is soldered. For more on this, see this answer of mine to an earlier question of yours.

According to ACuriousMind, physicists and mathematicians indeed use the words "diffeomorphism" and "isometry" in incompatible ways. I'm very surprised that I've never heard this before, because it seems like an important point to mention. Moreover, it's quite confusing because in both usages an "isometry" is a special case of a "diffeomorphism", but they are "offset": a mathematician's "isometry" is a physicist's "diffeomorphism". Since ACuriousMind gave some of this information in comments, for permanence here's a table translating between the two usages: $$\begin{array}{ccc} \text{Mathematicians' usage} & & \text{Physicists' usage} \\ \hline \text{diffeomorphism} & & \\[5pt] & \text{contains} & \\[10pt] \text{isometry} & = & \text{diffeomorphism} \\[5pt] & \text{contains} & \\[10pt] \text{autoisometry} & = & \text{isometry} \end{array}$$

Note that in your second quote from Carroll, he explicitly says that the metric and the matter fields are being transformed. Therefore a diffeomorphism, by his definition, is nothing more than a renaming of the points. It may be unnecessarily confusing that he talks about a map from manifold M to itself. If we're defining a manifold in the usual way, by point-set topology, then we can actually have manifolds that are different, in the sense that the set of points is different, but homeomorphic as manifolds. Then if we talk about a diffeomorphism from M to N, it becomes a little more clear that we have to transform the metric as well -- otherwise the metric would be a function that doesn't even have the right domain to operate on N.

Moreover, as discussed here, general diffeomorphisms do not map geodesics [...] to geodesics - only isometries do.

By Carroll's definitions, diffeomorphisms do map geodesics to geodesics. For example, suppose we do a diffeomorphism of the plane from Cartesian coordinates to polar coordinates. A line that is a geodesic under the metric $ds^2=dx^2+dy^2$ is also a geodesic under the metric $ds^2=dr^2+r^2d\theta^2$, which is what you get when you transform the metric according to the diffeomorphism. It's not a geodesic under the metric $ds^2=dr^2+d\theta^2$, which is what you get if you don't transform the metric and assume there is some natural correspondence between a point $(x,y)$ and a point $(r,\theta)$.