Yes. $M$ can be finitely generated.
Let $M=\mathbb{C}^{2 \times 2}$ denote $2 \times 2$ complex matrices. Let $A = \mathbb{C}^{2 \times 2} \oplus \mathbb{C}$ (a 5 dimensional complex vector space). Give $A$ the multiplication: $(P,s)(Q,t)=(PQ,st)$. Then your identity is $(I_2,1)$.
We see that $M$ is an ideal of $A$ (not just left but actually 2-sided).
Let $m\in\mathbb{C}$ and
$L_m = \left\{ \begin{pmatrix} a & ma \\ b & mb \end{pmatrix} \,{\Huge|}\, a,b \in \mathbb{C} \right\}$
This is certainly an uncountable set of left ideals (these matrices are preserved under row operations). Also, $\mathrm{dim}(L_m)=2$ so $\mathrm{dim}(M/L_m)=4-2=2$.
Notice that $L_m \cap L_n = \{ 0_{2 \times 2} \}$ [Note: You can think of $L_m$ sort of like lines through the origin.]
Everything is finitely generated.