There is a 2x2 matrix counter-example.

$$ M = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, \quad S_1 = \begin{pmatrix} 1 & 2 \\ 1 & 4 \end{pmatrix}, S_2 = \begin{pmatrix} a & 2c \\ c & d \end{pmatrix}. $$ Then $$ MS_1 = \begin{pmatrix} 1 & 2 \\ 2 & 8 \end{pmatrix}, \quad MS_2 = \begin{pmatrix} a & 2c \\ 2c & 2d \end{pmatrix} $$ so $MS_1$ is self-adjoint, and $MS_2$ is when $a,c, d\in\mathbb R$. Then $$ S_1S_2 = \begin{pmatrix} a + 2c & 2c+2d \\ a+4c & 2c+4d \end{pmatrix}, \quad S_2S_1 = \begin{pmatrix} a+2c & 2a+8c \\ c+d & 2c+4d \end{pmatrix}. $$ So these commute when $d = a + 3c$. But then $$ MS_2 = \begin{pmatrix} a & 2c \\ 2c & 2a+6c \end{pmatrix} $$ Tediously multiplying out shows these commute only when $2a=b$.

There is a 2x2 matrix counter-example.

$$ M = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, \quad S_1 = \begin{pmatrix} 1 & 2 \\ 1 & 4 \end{pmatrix}, S_2 = \begin{pmatrix} a & 2c \\ c & d \end{pmatrix}. $$ Then $$ MS_1 = \begin{pmatrix} 1 & 2 \\ 2 & 8 \end{pmatrix}, \quad MS_2 = \begin{pmatrix} a & 2c \\ 2c & 2d \end{pmatrix} $$ so $MS_1$ is self-adjoint, and $MS_2$ is when $a,c, d\in\mathbb R$. Then $$ S_1S_2 = \begin{pmatrix} a + 2c & 2c+2d \\ a+4c & 2c+4d \end{pmatrix}, \quad S_2S_1 = \begin{pmatrix} a+2c & 2a+8c \\ c+d & 2c+4d \end{pmatrix}. $$ So these commute when $d = a + 3c$. But then $$ MS_2 = \begin{pmatrix} a & 2c \\ 2c & 2a+6c \end{pmatrix} $$ Tediously multiplying out shows these commute only when $a=c$.