Yes. The possibility of staying at $M$ is irrelevant, so let's ignore it, so that the probability of an increase is 4/9 and the probability of a decrease is 5/9. For the moment, let's also ignore the actual sizes of these increases and decreases and let $X_n$ equal the total number of increases in the first n steps minus the total number of decreases in those steps. Then $X_n$ is a simple random walk on $\mathbb{Z}$ starting at 0 with probaility 4/9 of moving right and 5/9 of moving left. By a standard result on random walks on $\mathbb{Z}$, $X_n$ almost surely tends to $-\infty$. So for any k, we almost surely come to a point when we've had k more decreases than increases.

Going back to our original walk, every increase is by a factor of at most 2 and every decrease is by a factor of at least 2. So if we started at M and have had l increases and l+k decreases, then our current position is at most $2^l (1/2)^{l+k} M=(1/2)^k M$. So for k large enough, we must be at 0.