Suppose $M:[0, \infty)\to [0, \infty)$ is convex, non-decreasing, $M(0)=0$, $M'(0)=0$ (where the derivative is the derivative from the right), and $M(s)>0$ for all $s>0$. Under what conditions can we say that $$\sup \Bigl\{\frac{M(t)M(s/t)}{M(s)}: 0<s\leqslant t\leqslant 1\Bigr\}<\infty?$$
Under what conditions can we say there exists $ \Delta>0$ such that $$\sup\Bigl\{\frac{M(t)M(s/t)}{M(s)}: 0<t\leqslant 1, 0<s\leqslant \Delta t\Bigr\}<\infty$$
Let us address the first question. The condition \begin{equation} \sup \Big\{\frac{M(t)M(s/t)}{M(s)}: 0<s\le t\le 1\Big\}<\infty \tag{0} \end{equation} can be rewritten as follows: $M(t)M(v)\le CM(tv)$ for some real $C>0$ and all $t$ and $v$ in $(0,1]$. Letting $N(t):=M(t)/C[>0]$, this can be further rewritten as \begin{equation} N(t)N(v)\le N(tv) \tag{1} \end{equation} for $t$ and $v$ in $(0,1]$. For this (super)multiplicativity condition, such a "(super)additive" property as the convexity of $M$ (or equivalently, of $N$) seems of little, if any, help/relevance.
However, letting \begin{equation} n(s):=\ln N(e^{-s}) \tag{2} \end{equation} for real $s\ge0$, we can rewrite (1) in the (super)additive form: \begin{equation} n(r)+n(s)\le n(r+s)\tag{3} \end{equation} for real $r,s\ge0$. Also, (1) implies $N(1)\le1$, that is, $n(0)\le0$. So, if the function $n$ is convex, then $n(r)+n(s)\le n(r+s)+n(0)\le n(r+s)$ for $r,s\ge0$, so that (3) follows.
Thus, condition (0) will hold if $M$ is a positive multiple of a positive function $N$ on $[0,\infty)$ such that $N(1)\le1$ and the function $n$ defined by (2) is convex.
In particular, if $M(t)=Ct^b$ for positive $C$ and $b$, then (0) obviously holds and, on the other hand, for $N(t)=M(t)/C=t^b$ we have $n(s)=-bs$, which is convex in $s$.
Just want to add a cautionary note to Iosif Pinelis' answer. The convexity of $n$ is sufficient but not necessary for the super-additive condition $n(r)+n(s)\le n(r+s)$. An example where the necessity breaks down is $$n(t)=-te^{-t},$$ or $n(t)=t(1-e^{-t})$ if we need $n$ to be nonnegative.
