Information:
a-) $X$ and $Y$ are two continuous random variables on $\mathbb{R}$ having continuous distribution functions $F$ and $G$ with $G(y)\geq F(y)$ for all $y$.
b-) $S^X_n=\sum_{i=1}^n X_i$, $S^Y_n=\sum_{i=1}^n Y_i$, $A>0$, and $B<0$; where $X_i$ and $Y_i$ are i.i.d. replicas of $X$ and $Y$ respectively.
c-) $E[X]<0$ and $E[Y]<0$.
What I know:
By coupling (since $G\geq F$), I know that there exist a pair of random variables $(X^{'},Y^{'})$ such that $X=X^{'}$ in distribution, $Y=Y^{'}$ in distribution, and $X^{'}\geq Y^{'}$ almost surely. Using this result, I also have $S_n^{X^{'}}=\sum_{i=1}^n X^{'}_i\geq \sum_{i=1}^n Y^{'}_i=S_n^{Y^{'}}$. Since this holds for all $n$, I am able to compare the following:
$$\tau_A^{X^{'}}=\inf\{n\geq 0:S_n^{X^{'}}\geq A\}$$ $$\tau_A^{Y^{'}}=\inf\{n\geq 0:S_n^{Y^{'}}\geq A\}$$ $$\tau_B^{X^{'}}=\inf\{n\geq 0:S_n^{X^{'}}\leq B\}$$ $$\tau_B^{Y^{'}}=\inf\{n\geq 0:S_n^{Y^{'}}\leq B\}$$
with $\tau_A^{X^{'}}\leq \tau_A^{Y^{'}}$ since $S_n^{X^{'}}\geq A$ implies $S_n^{Y^{'}}\geq A$ and similarly $\tau_B^{X^{'}}\geq \tau_B^{Y^{'}}$. Please see (for details).
Claim-$1$:
$$E[\min\{\tau_A^{X^{'}},\tau_B^{X^{'}}\}]\geq E[\min\{\tau_A^{Y^{'}},\tau_B^{Y^{'}}\}]$$
holds for any $(A,B)$ and $(X,Y)$.
Claim-$2$:
$$E[\min\{\tau_A^{X^{'}},\tau_B^{X^{'}}\}]\geq E[\min\{\tau_A^{Y^{'}},\tau_B^{Y^{'}}\}]$$
holds for any $(A,B)$ and $(X,Y)$, if additionally $\partial F/\partial G$ is increasing.
I think that Claim 1 is not true in general. The following example is for discrete valued variables.
Take $X=\mathrm{Bernoulli}(\frac{1}{2})-\frac{5}{8}$ and $Y=\mathrm{Bernoulli}(\frac{1}{2})-\frac{7}{8}$. Then, $E[X]=-\frac{1}{8}$ and $E[Y]=-\frac{3}{8}$. Take $A=\frac{1}{4}$ and $B=-\frac{1}{2}$. Then, $\min\{\tau_A^{X^{'}},\tau_B^{X^{'}}\}=1$ and $P[\min\{\tau_A^{Y^{'}},\tau_B^{Y^{'}}\}>1]\geq\frac{1}{2}$. Note that $E[\min\{\tau_A^{X^{'}},\tau_B^{X^{'}}\}]=1$ and $E[\min\{\tau_A^{X^{'}},\tau_B^{X^{'}}\}]\geq \frac{3}{2}$.
But the continuous one can be constructed by taking approximation: Let $Z$ be a gaussian variable of mean $0$ and variance $1$. For $\epsilon>0$, define $X^{(\epsilon)}=X+\epsilon Z$ and $Y^{(\epsilon)}=Y+\epsilon Z$. Then, $X,Y$ satisfy the condition in Claim 1. We will show that for $A=\frac{1}{4}$ and $B=-\frac{1}{2}$, Claim 1 is violated for small enough $\epsilon$. Roughly speaking, we would like to say that small perturbation $\epsilon Z$ just increases $E[\min\{\tau_A^{X^{'}},\tau_B^{X^{'}}\}]$ a little. Moreover, the small perturbation $\epsilon Z$ cannot decrease $E[\min\{\tau_A^{Y^{'}},\tau_B^{Y^{'}}\}]$ too much. To be more precise, let $(X_i)_i$ be i.i.d copies of $X$ and $(Z_i)_i$ be i.i.d copies of $Z$. Denote by $S^{X^{(\epsilon)}}_k$ the partial sum of $X_i^{(\epsilon)}=X_i+\epsilon Z_i$. Then, $$P[S^{X^{(\epsilon)}}_k\geq B]\leq P\left[X_1+\cdots+X_k-kE[X]\geq \frac{k}{16}\right]+P\left[\epsilon(Z_1+\cdots+Z_k)+kE[X]\geq -\frac{k}{16}-\frac{1}{2}\right].$$ Recall that $E[X]=-\frac{1}{8}$. By Hoeffding's inequality, $$P\left[X_1+\cdots+X_k-kE[X]\geq \frac{k}{16}\right]\leq \exp\left(-\frac{k}{128}\right).$$ Since $\epsilon(Z_1+\cdots+Z_k)$ is a mean zero gaussian variable with variance $k\epsilon^2$, $$P\left[\epsilon(Z_1+\cdots+Z_k)+kE[X]\geq -\frac{k}{16}-\frac{1}{2}\right]\leq P[\mathcal{N}(0,1)\geq \frac{k-8}{16\sqrt{k}\epsilon}].$$ Then, there exists a universal constant $C$ such that $\sup\limits_{0<\epsilon\leq 1}E[\tau^{X^{(\epsilon)}}_{B},\tau^{X^{(\epsilon)}}_B>k]\leq \frac{C}{k}$. (Note that $P[\tau^{X^{(\epsilon)}}\geq k+1]\leq P[S^{X^{(\epsilon)}}_k\geq B]$.) In other words, the tail of $\min(\tau^{X^{(\epsilon)}}_{B},\tau^{X^{(\epsilon)}}_{A})$ is uniformly small. Pick $k=K_0$ large enough such that $$E[\min(\tau^{X^{(\epsilon)}}_{B},\tau^{X^{(\epsilon)}}_{A}),\min(\tau^{X^{(\epsilon)}}_{B},\tau^{X^{(\epsilon)}}_{A})\geq K_0]\leq 0.01.$$ On the other hand, for $i=2,\ldots,K_0-1$, $$P[\min(\tau^{X^{(\epsilon)}}_{B},\tau^{X^{(\epsilon)}}_{A})=i]\leq P[\min(\tau^{X^{(\epsilon)}}_{B},\tau^{X^{(\epsilon)}}_{A})>1]\leq P[\epsilon |Z_1|>1/4].$$ Thus, $$E[\min(\tau^{X^{(\epsilon)}}_{B},\tau^{X^{(\epsilon)}}_{A}),1<\min(\tau^{X^{(\epsilon)}}_{B},\tau^{X^{(\epsilon)}}_{A})<K_0]\leq (K_0)^2P[\epsilon |Z_1|>1/4].$$ Then, we can pick a small enough $\epsilon$ such that $$E[\min(\tau^{X^{(\epsilon)}}_{B},\tau^{X^{(\epsilon)}}_{A})]\leq 1+0.02.$$ By a simpler argument as in the last step, $$E[\min(\tau^{Y^{(\epsilon)}}_{B},\tau^{Y^{(\epsilon)}}_{A})]\geq 1.5-0.01.$$