Is there anything noteworthy about the following set?
$$ \Omega_k ≔ \left\{e^A \mid \operatorname{rank}(A)=k \right\} $$
Are there any special algebraic properties to $e^A$ in this case? Does the manifold have any significant properties? What's its dimensionality?
Obviously $Ω_0 = \{\}$. In the case $k=1$, we have $A = u v^\top$ for some vectors $u,v$, hence
$$ \Omega_1 = \Big\{+\frac{e^{u^⊤v}-1}{u^\top v} uv^\top \mid u, v \in \Bbb R^n, u v^\top \neq 0 \Big\} $$
So $Ω_1$ consists of a subset of the set of rank-1 perturbations of the identity matrix. Note that $\frac{e^{u^⊤v}-1}{u^⊤v}$ is a shorthand for $φ_1(u^⊤v)=∑_{k∈ℕ} \frac{(u^⊤v)^k}{(k+1)!}$ which is defined even when $u⟂v$.
Since matrix exponentiation commutes with conjugation, and because of the rank-nullity theorem, we can reduce the problem to understanding the image of Jordan blocks.
If $N$ is nilpotent, then $N$ and $kI$ commute, so $e^{kI+N}=e^{kI}e^N=e^ke^N$, and so we can further reduce to understanding $e^N$.
Lemma: if $N^m\neq 0$ but $N^{m+1}=0$, then $(e^{N}-I)^m\neq 0$ but $(e^{N}-I)^{m+1}= 0$
Proof: $e^N-I=N(I+N/2+N^2/6+\ldots)$, so $(e^N-I)^{\ell}$ is a a multiple of $N^{\ell}$, which is zero if $\ell>m$. On the other hand, fully expanding out the power series and using $N^{m+1}=0$ we get $(e^N-I)^m=N^m\neq 0$.
With this, one can show that if $J$ is a single Jordan block, then $e^J$ is conjugate to a single Jordan block.
As as consequence of this observation, one can show that $\ker(e^A-I)$ is the direct sum of the eigenspaces of $A$ with eigenvalue $2\pi i k, k\in \mathbb Z$. Further, every matrix $B$ such that $B$ is invertible and $\dim\ker B-I \geq n-k$ can be written as $e^A$ for some $A$ with rank $k$.
Finally, appealing to the rank nullity theorem, we conclude
$$\Omega_k=\{B\in GL_n(\mathbb C) \mid \operatorname{rank}(B-I)\leq k\}$$
With this final result in mind, there is actually a simple argument to get half of it:
Because $\operatorname{im}(AB)\subset \operatorname{im}(A)$, we have that $\operatorname{im}(e^A-I)=\operatorname{im}(A(I+A/2+A^2/6+\ldots))\subset \operatorname{im}(A)$, and so the rank is non-increasing. This gives us inclusion one way.
Unfortunately, I don't yet see a way to get the other inclusion without something akin to JNF.