$\newcommand\ss{\text{ss}}\newcommand\Z{\mathbb Z}$For real $t\ge0$, let $X_t:=mt+W_t$, where $m$ is a real number and $W_\cdot$ is a standard Brownian motion. So, $X_\cdot$ is a drifted Brownian motion starting at $0$ with the constant drift coefficient $m$.
For real $c$, let $T_c:=\min\{t\ge0\colon X_t=c\}$.
According to formula 3.0.2 on p. 309 in Handbook of Brownian Motion - Facts and Formulae (Second Edition) by A. Borodin and P. Salminen, the pdf, say $p_{a,b}$, of the random variable (r.v.) $T_{a,b}:=\min(T_a,T_b)$ is given by the formula
$$p_{a,b}(u)=e^{-m^2u/2}\big(e^{ma}\ss_u(b,b-a)+e^{mb}\ss_u(-a,b-a)\big)$$
for real $u\ge0$,
where $-\infty<a<0<b<\infty$ and, according to the notation list on p. 641 of the mentioned handbook,
$$\ss_u(c,d):=\sum_{k\in\Z}\frac{d-c+2k d}{u^{3/2}\sqrt{2\pi}}
\exp\Big\{-\frac{(d-c+2k d)^2}{2u}\Big\}.$$
So, for real $t\ge0$,
$$Q_{a,b}(t):=P(T_{a,b}>t)=\int_t^\infty du\,p_{a,b}(u). \tag{1}$$
Thus, the probability in question is $$P(T_a\le t,T_b\le t)=1+P(T_a>t,T_b>t)-P(T_a>t)-P(T_b>t) \\ =1+Q_{a,b}(t)-Q_{a,\infty-}(t)-Q_{-\infty+,b}(t),$$ with $Q_{a,b}(t)$ given by (1).
Using e.g. formula 2.0.2 on p. 295 of the mentioned handbook, the expressions for $P(T_a>t)=Q_{a,\infty-}(t)$ and $P(T_b>t)=Q_{-\infty+,b}(t)$ can be simplified as follows: $$P(T_z>t) =\frac1{2z} \Big(1-e^{2 m z}-\text{erf}\left(\frac{m t-z}{\sqrt{2t}}\right)+e^{2 m z} \text{erf}\left(\frac{m t+z}{\sqrt{2t}}\right)\Big)$$ for $z\in\{a,b\}$.