No. It is not an equivalence. We need both the properties to be satisfied.
Given any matrix $M$, and any vector $v$ and a scalar $\sigma$, we can get a $u$, if we just want to satisfy $Mv = \sigma u.$ This need not satisfy $M^*u = \sigma v$
For instance, let $M = \begin{bmatrix} 2 & 9\\ 3 & 6 \end{bmatrix}$, $v = \begin{bmatrix}1 \\ 0 \end{bmatrix}$ and $\sigma = 2$. This gives us $Mv = \begin{bmatrix} 2\\ 3 \end{bmatrix}$ and hence $u = \begin{bmatrix} 1 \\ 3/2 \end{bmatrix}$. However, $\dfrac1{\sigma} M^*u = \dfrac12 \begin{bmatrix} 2 & 3\\ 9 & 6 \end{bmatrix} \begin{bmatrix} 1 \\ 3/2 \end{bmatrix} = \dfrac12 \begin{bmatrix} 2 + 9/2 \\ 9 + 9 \end{bmatrix} = \dfrac12 \begin{bmatrix}11/2 \\ 18 \end{bmatrix} = \begin{bmatrix}11/4 \\ 9 \end{bmatrix} \neq v$
In fact, we see that we need to choose $v$ to be the eigenvector of $M^*M$ and choose $u$ to be the eigenvector of $MM^*$ to get what we want.