What you want is true if and only if there exists another very ample line bundle $\scr L'$ on $M$ such that $H^0(M, \, \scr L' \otimes L^{-1})=\{0\}$.

In fact, this condition is equivalent to say that no linear section in the embedding provided by $\scr L'$ splits as $S_M$ plus an effective divisor, that is precisely your request.

In particular, the answer is negative when the ample cone of $M$ is generated by $\scr L$, as explained in abx's comment.

For an example in the positive direction, take as $M$ the Hirzebruch surface $\mathbb{F}_e$, denote by $C_0$ the unique section of negative self-intersection and by $\mathfrak f$ the fibre, and set $$\mathscr{L} = C_0 + n \mathfrak{f}, \quad \mathscr{L}' = C_0 + m \mathfrak{f},$$
with $n > m > e$ (as usual, I do not distinguish between divisors and line bundles).

Then both $\scr L$ and $\scr L'$ are very ample by [Hartshorne, *Algebraic Geometry*, Theorem 2.17 p. 379], and we have
$$H^0(\mathbb{F}_e, \, \mathscr{L}' \otimes \mathscr{L}^{-1}) = H^0(\mathbb{F}_e, \, (m-n) \mathfrak{f})=\{0\},$$ because $m-n <0$.

**Edit.** The "only if" part actually holds if we consider embeddings induced by *complete* linear systems, see C. McMullen answer below.