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].
If $R_x$ rotates around the $x$-axis, and $R_y$ rotates around the $y$-axis, and you want to rotate first around $x$, and then around $y$, simply apply $R_y R_x$ to your vector, let's call it $v$.
This is because $R_x v$ rotates $v$ around the $x$-axis, then $R_y(R_x v)$ rotates $R_x v$ around the $y$-axis.
Best Answer
When composing two rotations, it is useful to know that a rotation about $\alpha$ about an axis $\ell$ can be written as the composition of two reflections in planes containing $\ell$, the first being chosen arbitrarily and the second being at an (oriented) angle $\frac\alpha2$ with respect to the first. Now in the composition of $4$ reflections you get, you can make your choices so that the second and third planes of reflection (the second reflection for the first rotation and the first reflection for the second rotation) are both equal to the unique plane passing through the two axes. Then poof!, those second and third reflections annihilate each other, and you are left with the composition of the first and the fourth reflection, which is a rotation with axis the intersection of those planes, and angle twice the angle between those planes.
If you want to calculate the axis and angle in terms of the original angles, formulas get a bit complicated (even for very easy choices of initial axes as in the question), but such is life, the concrete answer isn't really very easy to write down or remember.