Can the total amount of solution be found as a ratio between molar mass of a component and total mass of solution?
To sum up the comments, only the following relation for the total amount of solution $n_\mathrm{tot}$ is universally true:
$$n_\mathrm{tot} = n_\mathrm{solvent} + n_\mathrm{solute} = \frac{m_\mathrm{solvent}}{M_\mathrm{solvent}} + \frac{m_\mathrm{solute}}{M_\mathrm{solute}}\tag{1}$$
The best you can do is to assume that $n_\mathrm{tot}\approx n_\mathrm{solvent}$ for the diluted solutions of small molecules. Also, if the molar masses are similar $(M_\mathrm{solvent}\approx M_\mathrm{solute}\approx \bar{M}),$ the expression can be lead to a common denominator:
$$n_\mathrm{tot} \approx \frac{m_\mathrm{solvent} + m_\mathrm{solute}}{\bar{M}}\tag{2}$$
This can be the case, for example, for the solution of ammonium nitrate $(M(\ce{NH4NO3}) = \pu{80.043 g mol^-1})$ in dimethyl sulfoxide $(M(\text{DMSO}) = \pu{78.13 g mol^-1}).$
Algebra aside, your mistake was also neglecting $M_\mathrm{solvent}.$
Keep in mind you can always check dimensions.
For your proposed formula

$$n_\mathrm{solvent} + n_\mathrm{solute} = \frac{M}{m_\mathrm{solvent} + m_\mathrm{solute}}$$

it doesn't work out well:
$$\mathrm{dim}~n_\mathrm{tot} = \mathsf{M}\cdot\mathsf{N}^{-1}\cdot\mathsf{M}^{-1} = \mathsf{N}^{-1} \neq \mathsf{N}\tag{3}$$
Illustrating with common units used in chemistry:
$$[n_\mathrm{tot}] = \frac{[M]}{[m_\mathrm{solvent}] + [m_\mathrm{solute}]} = \frac{\pu{g mol^-1}}{\pu{g}} = \pu{mol^-1} \neq \pu{mol}\tag{4}$$