Let $A$ be an arbitrary matrix in $\mathbb C$, and write its SVD as $A=UDV^\dagger$.
One can show that $A$ being normal is equivalent to the eigenspaces of $A^\dagger A$ and $AA^\dagger$ being equal (because $AA^\dagger=A^\dagger A$ if and only $UD^2 U^\dagger=VD^2 V^\dagger$, which then translates into a condition on the eigenspaces).
Can we also tell whether $A$ is diagonalisable, working directly from its SVD?
Of course, from the SVD one also knows the matrix itself, and then any diagonalisability criterion can be used. However, I'm asking specifically about whether there is a way to infer the diagonalisability of $A$ directly from the structure of its left and right principal components, as we can do for its normality.
For example, working in dyadic notation, if $e_0$ and $e_1$ are two orthogonal vectors, and $e_+\equiv e_0+e_1$, then $e_0 e_1^*$ is not diagonalisable while $e_+e_0^*$ and $e_+e_1^*$ are. More generally, we have that $v w^*$ is diagonalisable if and only if $v$ and $w$ are not orthogonal, which follows easily from . $$(v w^*)^2=\langle w,v\rangle (vw^*)$$ meaning that $vw^*$ has minimal polynomial $x\mapsto x(x-\langle w,v\rangle)$ and is thus diagonalisable iff $\langle w,v\rangle\neq0$.
In matrix notation, this is saying that a matrix $A$ that satisfies $Av=w$ for some $w$, and is zero on every other vector, is diagonalisable if and only if $\langle v,w\rangle\neq0$.
Is there a similar argument for more general matrices which are not simply writable as a single dyad (that is, whose range is more than one-dimensional)?
This at least is impossible for non-normal $2\times2$ matrices, if we only know their singular vectors but not their singular values. (Note that in your example of a rank-$1$ matrix $A$, some information about the singular values of $A$ is known, namely, $A$ has a zero singular value.)
It is not hard to show that if $U$ is a $2\times2$ unitary matrix that is not diagonal, there always exist two nonnegative diagonal matrices $S_1$ and $S_2$ such that $US_1$ is diagonalisable but not normal and $US_2$ is not diagonalisable.
It follows that when $A$ is $2\times2$ and not normal, we can always construct a non-normal matrix $B$ that shares the same singular vectors with $A$, and we can make $B$ diagonalisable or otherwise at will by tuning its singular values. Consequently, under the hypothesis that $A$ is a $2\times2$ non-normal matrix, we can never determine its diagonalisability solely based on its singular vectors.