Yes, this is true, and you are right, this follows from a generalization of the Brunn-Minkowski inequality.

Let $K_s = \{x \mid f(x) \le s\}$, so that $M_s = \partial K_s$. We have $K_s \supseteq (1-s)K_0 + sK_1$, thus the surface area of the former is $\ge$ the surface area of the latter.

The surface area of a convex body can be written as a mixed volume:
$$
\mathrm{vol}_{n-1}(\partial K) = n \mathrm{vol}(\underbrace{K, \ldots, K}_{n-1}, B).
$$
A general version of the Brunn-Minkowski inequality says that the function
$$
\mathrm{vol}(\underbrace{(1-t)K_0 + tK_1, \ldots, (1-t)K_0 + tK_1}_{n-i}, L_1, \ldots, L_i)^{\frac1{n-i}}
$$
is concave for any convex bodies $L_1, \ldots, L_i$.

It follows that the $(n-1)$-st root of the surface area of $K_t$ is concave. Concavity of $f^{\frac{1}{n-1}}$ implies concavity of $\log f$, and we are done.

References:

*Gardner, R. J.*, **The Brunn-Minkowski inequality**, Bull. Am. Math. Soc., New Ser. 39, No. 3, 355-405 (2002). ZBL1019.26008, Section 17.

*Burago, Yu. D.; Zalgaller, V. A.*, Geometric inequalities. Transl. from the Russian by A. B. Sossinsky, Grundlehren der Mathematischen Wissenschaften, 285. Berlin etc.: Springer-Verlag. XIV, 331 p.; DM 184,- (1988). ZBL0633.53002, p. 146.

*Schneider, Rolf*, Convex bodies: the Brunn-Minkowski theory, Encyclopedia of Mathematics and Its Applications. 44. Cambridge: Cambridge University Press. xiii, 490 p. (1993). ZBL0798.52001, Theorem 6.4.3.