The answer is yes, since the exponential map $\exp: \mathbf{so}(3) \rightarrow \mathbf{SO}(3)$ is surjective (=onto).
Long answer:
Axis-angle can be represented using a $3$-vector $\omega$ while the magnitude $\theta=|\omega|$ is the rotation angle and $\mathbf{u}=^\omega/_\theta$ is the rotation axis. 3-Vectors are closed under the cross product:
$$\omega_1\in \mathbb{R}^3, \omega_2\in \mathbb{R}^3\Rightarrow (\omega_1\times \omega_2)\in\mathbb{R}^3.$$
Each such vector $\omega$ has an equivalent $3\times 3$ matrix representation
$\hat{\omega}$ (which is uniquely defined by $\hat{\omega}\cdot \mathbf{a} := \omega\times \mathbf{a}$ for $\mathbf{a}$ being a general 3-vector).
The space of matrices of the form $\hat{\omega}$ is called the Lie algebra $\mathbf{so}(3)$. Thus, one can show that matrices of the form $\hat{\omega}$ are closed under the Lie bracket $[A,B]=AB-BA$:
$$\hat{\omega_1}\in \mathbf{so}(3), \hat{\omega_2}\in \mathbf{so}(3)\Rightarrow [\hat{\omega}_1, \hat{\omega}_2]\in\mathbf{so}(3).$$
Now, let us consider the matrix exponential: $\exp(\mathtt{A}) = \sum_{i=0}^\infty \frac{\mathtt{A}^i}{i!} $. Two poperties can be shown:
(1) If $\hat{\omega}\in\mathbf{so}(3)$, then $\exp(\hat{\omega})\in\mathbf{SO}(3)$.
$\mathbf{SO}(3)$ is the special orthogonal group in three dimensions. Thus, it consists of matrices which are orthogonal ($\mathtt{R}\cdot \mathtt{R}^\top=\mathtt{I}$) and the determinant is 1. In other word, it is the group of pure rotations.
(2) The exponential map $\exp: \mathbf{so}(3) \rightarrow \mathbf{SO}(3)$ is surjective.
So, (1) says that every $\exp(\hat{\omega})$ is a rotation matrix. And, (2) says that for each rotation matrix $\mathtt{R}$, there is at least one axis-angle
representation $\omega$ so that $\exp(\hat{\omega})=\mathtt{R}$
Proofs of (1) and (2) are in corresponding text books, e.g. [Gallier, page 24].
Please see this diagram. Lines $m,n$ are normals to reflexive axes with the angle between them $\frac\theta2$. Then $v''$, which is reflected twice by $m,n$ is such a vector rotated $\theta$ from the original vector $v$.
And I think this has also an algebraic explanation in geometric algebra. The reflection of $v$ by the axis $n$ is represented as $v'=-nvn$. Then $v''=-mv'm=-m(-nvn)m=(mn)v(nm)=RvR^\dagger$, where $R=mn$ and $R^\dagger$ is reverse of $R$. We speak of $R$ is rotor of angle $\theta$ if $m\cdot n=\cos\frac\theta2$. $RvR^\dagger$ is exactly the expression of a rotation in geometric algebra.
Best Answer
I believe constructing the reflection planes and starting from the 2D analogue would help.
Let's say the rotation is about the origin, and is nontrivial.
Well, after the rotation happens, the origin is still at the origin.
Hence either both reflections change the position of the origin, or none of the reflections changes the position.
However, if both reflections change the position of the origin, and both reflections are not the same, then the origin will change position.
Hence neither of the reflections change the position of the origin, i.e. the line that the reflection is done in passes through the origin.
Let's bring this back to 3D.
Every point on the axis goes back to being the same point. By similar logic, the planes must pass through the axis.
The only variable that can now vary is the angle the pair of planes form. With only one variable, the rest should be easier.