Let $X, Y$ be non negative random variables with finite expectation. We say that $Y$ stochastically bounds $X$ if there exists some $C > 0$ such that for all $x \in \mathbb R$,
$$\mathbb P(X \geq x) \leq C \, \mathbb P(Y \geq x).$$
If $C$ can be taken equal to or less than $1$, we say that $Y$ stochastically dominates $X$.
Question: Suppose $Y$ stochastically bounds $X$. Does there exist some $C > 0$ such that $CY$ stochastically dominates $X$?
Not if $C>1$. (The case $C=1$ is trivial, and $C$ cannot be $<1$.)
Indeed, let $H(x):=P(Y\ge x)$ and $G(x):=\min(1,CH(x))$ for real $x$. Then $G(x)=P(X\ge x)$ for some random variable $X$ and all real $x$ (because $G$ is a left-continuous function decreasing, non-strictly, from $1$ to $0$ on $\Bbb R$). Moreover, the condition $P(X\ge x)\le CP(Y\ge x)$ obviously holds for all real $x$.
If $KY$ stochastically dominates $X$ for some real $K>0$, then $H(x/K)=P(KY\ge x)\ge P(X\ge x)=G(x)=\min(1,CH(x))$ for all real $x$. But this cannot be true for $x$ in some right neighborhood of $0$ if $H(0+)=1$ but $H(x)<1$ for $x>0$.
The OP asked if we can ensure $$P(X\ge x)\le P(KY\ge x)$$ for some $K>0$ and all large enough $x$. The answer here is also no, if $C>1$. Indeed, suppose that $$P(Y\ge x)=\frac1{\ln x}$$ for $x\ge e$ and $$P(X\ge x)=\min(1,CP(Y\ge x))$$ for all real $x$. Then the condition $P(X\ge x)\le CP(Y\ge x)$ obviously holds for all real $x$. However, for any $C>1$, any real $K>0$, and all large enough $x>0$, $$P(X\ge x)=CP(Y\ge x)=\frac C{\ln x}\not\le \frac1{\ln(x/K)}=P(KY\ge x).$$
This does not hold, at least if you want a constant independent of $Y$. This can be seen by letting $X$ be uniform on $I_0 :=[0,1]$, and $Y$ be uniform on $I_0\cup I_1$ for $I_1 := [D, D+1]$ with $D$ sufficiently large. Then, $Y$ stochastically bounds $X$ (with constant 2), but does not stochastically dominate $X$ (one would need to choose $C$ dependent on $Y$, specifically dependent on $D$).
