How can I find a vector when multiplied by singular matrix? I have the following equation:
$$Mv=MKu$$
Where $M$ is a $3\times3$ singular matrix, $K$ is a $3\times3$ invertible matrix, and $v$ is a known vector.
How can I find $u$?
If $M$ was invertible, I assume the solution would be $u:=K^{-1}v$.
Is there a solution if $M^{-1}$ doesn't exist?

Since $K$ is invertible, for a given $v$ the vector $u_1 = K^{-1}v$ solves the equation $$Mv = MKu_1,$$
as pointed out in the comments by Michael Burr and user254665: $MKu_0 = MKK^{-1}v = MIv = Mv$.
However, the choice of $u=u_1$ is not unique. Take any $v_0 \in \ker M$, i.e.  with $Mv_0 = 0$, and let $u_0 = K^{-1}v_0$   then we have for $u=u_0 + u_1$:
$$MKu = MKu_0 + MKu_1 = Mv_0 + Mv = 0 +  Mv.$$
Thus, the affine space $K^{-1}v + K^{-1}( \ker M)$ solves the equation. 
Coversely, suppose $u$ solves the equation. We'd like to show that $u-K^{-1}v \in K^{-1}\ker M$, or equivalently $Ku-v \in \ker M$. We have indeed $$M(Ku - v)=MKu - Mv = Mv -Mv = 0.$$
In conclusion, every solution is of the form $$u=K^{-1}v + K^{-1}v_0$$ for some $v_0\in \ker M$. So if you are looking for a specific $u$, you're at least free in the choice of $v_0$.
