If we have the following eigenvalues and eigenvectors $$A \psi = a\psi$$ and $$B\psi = b\psi$$ we can easily show that they have same eigenvector if they commute.
$$AB\psi=Ab\psi=bA\psi=ba\psi$$ $$BA\psi=Ba\psi=aB\psi=ab\psi$$ So they commute! They satisfy the proposal. $$[A,B]=0$$
But what if our eigenvectors are different instead of being same with corresponding eigenvalues. Can we show it again so that they have same eigenvector?
$$A\alpha_n = a_n\alpha_n$$ $$B\beta_n = b_n\beta_n$$
On
If two operators, $\hat{A}$ and $\hat{B}$, share the same orthonormal basis, $|j\rangle$ and can be decomposed by the spectral theorem, it is fairly easy to show that the operators also commute.
We can express the eigenvalues of the decomposition as, $$ \begin{align} \hat{A}|j\rangle &=a_j|j\rangle \\ \hat{B}|j\rangle &= b_j|j\rangle. \end{align} $$
Thus the matrices when multiplied can be expressed as, $$ \begin{align} \hat{A} \hat{B} &= \sum_{j,k} a_j |j\rangle \langle j| b_k |k\rangle \langle k| \\ & = \sum_{j,k} a_j b_k |j\rangle \langle j|k \rangle \langle k| \\ &= \sum_{j,k} a_j b_k \delta_{j,k} |j\rangle \langle k| \\ &= \sum_j a_j b_j |j\rangle \langle j| \end{align} $$
Since the component eigenvalues values commute, so too do the operators. $$ \begin{align} \left[\hat{A},\hat{B}\right] & = \sum_j a_j b_j |j\rangle \langle j| - \sum_j b_j a_j |j\rangle \langle j| \\ & = \sum_j \left(a_j b_j - b_j a_j\right) |j\rangle \langle j| = 0 \end{align} $$
(Edit)
To answer your question, they will not commute unless they share the same basis. You can see this in the third line of my derivation of the operator multiplication. The expression $\langle j | k\rangle = \delta_{j,k}$ only holds if the operators share the same basis. If they don’t then you have to do standard matrix multiplication which will definitely not commute.
I'm this as the following concern:
There is a theorem which says :
If $\Omega$ and $\Lambda$ are two commuting Hermitian operators, there exists (at least) a basis of common eigenvectors that diagonalizes them both.
The main concern is a complete set of eigenbasis. Note the operator should be Hermitian so that there exists at least a basis of its orthonormal eigenvectors.
The proof for theorem in case of nondegenerate given by:
$$\Omega|\omega_i\rangle=\omega_i|\omega_i\rangle$$ $$\Lambda\Omega|\omega_i\rangle=\Lambda\omega_i|\omega_i\rangle$$ $$\Omega\Lambda|\omega_i\rangle=\omega_i\Lambda|\omega_i\rangle$$ i.e. $\Lambda|\omega_i\rangle$ is an eigenvector of $\Omega $ with eigenvalue $\omega_i$. Since this vector is unique up to a scale $$\Lambda|\omega_i\rangle=\lambda_i|\omega_i\rangle$$ Thus $|\omega_i\rangle$ is also a eigenvector of $\Lambda$ with eigenvalue $\lambda_i$.