Let's focus our attention on a straightforward special case: the quotient map $\mathbb{R} \to \mathbb{R} / \mathbb{Z}$, as a map of Lie groups, induces an isomorphism on Lie algebras, but it is not an isomorphism. So the Lie algebra cannot even tell whether a Lie group is compact or not (although it can get surprisingly close, as it turns out, modulo this example).
When you try to reconstruct a connected Lie group as the group generated by the image of the exponential map, the problem is that there are some relations between these generators which take place "far from the identity," and without more global information than just the Taylor expansion of the Lie group multiplication at the identity, you don't know which of these relations to impose or not. In the above example, one such relation is $\frac{1}{2} + \frac{1}{2} = 0$, which holds in $\mathbb{R} / \mathbb{Z}$ but not in $\mathbb{R}$.
(What's worse, you don't even know a priori that you can consistently impose these relations in such a way that you get a Lie group globally. That is, it's not at all obvious, although it is true, that every finite-dimensional Lie algebra is the Lie algebra of a Lie group.)
(Personally I use $\mathrm{SU}(2)$ strictly to denote the group of $2\times 2$ special unitary matrices and $\mathrm{Sp}(1)$ to denote the set of unit quaternions, but most people use them interchangeably since they are isomorphic Lie groups, like $\mathrm{U}(1)$ and $\mathrm{SO}(2)$.)
Do both the set of imaginary quaternions and the set of unit quaternions parameterize SU(2) via the exponential map?
This is kind of like asking "do both pairs of angles and the sphere parametrize the sphere via sines and cosines?" It's a weird thing to say. One uses sines and cosines to parametrize the sphere using two angles, sure, but one uses ... the identity function to parametrize a space with itself.
The unit quaternions are $\mathrm{Sp}(1)$, and each unit quaternion has a polar form given by the formula$\exp(\theta\mathbf{u})=\cos(\theta)+\sin(\theta)\mathbf{u}\,$ (just like complex numbers) where $\mathbf{u}$ is unit vector. Thus, the exponential function maps $\mathrm{Im}(\mathbb{H})=\mathbb{R}^3$ to the group $\mathrm{Sp}(1)$.
I don't understand the rest of what you're saying in question (1).
Is it true that each of the following three sets constitutes a Lie group: The set of unit quaternions, the set of imaginary quaternions, and the set of all nonzero quaternions?
The unit quaternions and nonzero quaternions are multiplicative groups and the imaginary quaternions are a vector space under addition, but the imaginary quaternions are certainly not closed under multiplication. Given two vectors $\mathbf{u}$ and $\mathbf{v}$, the product's real and imaginary parts (i.e. its scalar and vector components) are given by the formula $\mathbf{uv}=-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$, so the (quaternion) product of two vectors will itself be a vector if and only if they're orthogonal.
It's not clear to me how to define the appropriate metric for this to be the case.
There's already a metric on $\mathbb{H}$. It's given by
$$ |x_0+x_1\mathbf{i}+x_2\mathbf{j}+x_3\mathbf{k}|=\sqrt{x_0^2+x_1^2+x_2^2+x_3^2}. $$
So if you have a nice curve $\gamma:[0,1]\to\mathbb{H}$ its length can be calculated by $\int_0^1 |\gamma'(t)|\,\mathrm{d}t$, just as in any Euclidean space $\mathbb{R}^n$. Indeed, just as in $\mathbb{R}^2$ and $\mathbb{R}^3$, the length of an arc between two points on the unit sphere will be the angle between them.
For the unit quaternions, the tangent space at $1$ is just the imaginary quaternions; for non-unit quaternions, isn't the tangent space the same?
Nope. For inspiration, just look at $S^2$. Clearly antipodal points have the same tangent plane (interpreted as vector subspaces of $\mathbb{R}^3$ anyway), but otherwise the various tangent planes are not parallel and exist in different orientations.
Suppose $q:[0,1]\to\mathrm{Sp}(1)$ is a differentiable curve in the unit quaternions, and that $q(0)=1$. We have $q(t)\overline{q(t)}=1$; differentiating and then evaluating at $t=0$ yields $q'(0)+\overline{q'(0)}=0$, which is only possible if $q'(0)$ is purely imaginary since $x+\overline{x}=2\mathrm{Re}(x)$. But this argument relies on $q(0)=1$, i.e. on the fact we're look at the tangent space at $q=1$.
In general if $q(0)=q$ then instead you get $q'(0)\overline{q}+q\overline{q'(0)}=0$, or equivalently $2\mathrm{Re}(q'(0)\overline{q})=0$, so that $q'(0)\overline{q}$ is pure imaginary, or in other words $q'(0)$ is an an imaginary quaternion multiplied by $q$ from the right. Since $q^{-1}\mathbf{x}q$ is just $\mathbf{x}$ rotates, and $\mathbf{x}q=q(q^{-1}\mathbf{x}q)$, we could have equally well have said multiplying by $q$ from the left instead of the right.
In conclusion, the act of multiplying by $q$ slides (is a map from) the tangent space $T_1S^3$ to the tangent space $T_qS^3$. (Note $S^3$ is another term for $\mathrm{Sp}(1)$, just as $S^1$ is notation for $\mathrm{U}(1)$.)
If you want the exponential map $\exp:T_qS^3\to S^3$ in the Riemannian manifold sense, you can do a "transport of structure" argument: slide $T_qS^3$ to $T_1S^3$, apply the usual $\exp$ map, then slide back (so that $\exp$ of the zero vector in $T_qS^3$ yields $q$). Depending on if you slide using left or right multiplications, this yields $q\exp(\overline{q}x)$ or $\exp(x\overline{q})q$ (which are equal to each other).
Hopefully this answers (4) as well for you.
Best Answer
You can really take your favorite nonabelian Lie group and any metric that isn't bi-invariant. The simplest example is $$ G = \left\{ \begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix} \mid a,b \in \mathbb{R} \right\} .$$
This group acts simply transitively by isometries on the hyperbolic plane $\mathbb{H}^2 = \{x+iy \mid y >0 \}$, and admits a (unique) left-invariant Riemannian metric making the orbit map an isometry. For that metric, the geodesics in $G$ correspond to the geodesics in the hyperbolic plane.
The one-parameter subgroups need not correspond to Riemannian geodesics. For example, the curve $$ c(t) = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix} \cdot i = t + i $$ is a horocycle.