I claim this is not an answer but some calculus following Thomas's observation. As I cannot write it as a comment, I write it down here. The stochastic integral

$$\left(M_t:=\int_0^t\frac{\sigma}{1+m(s)}dW_s\right)_{t\ge 0}$$

is a continuous martingale starting from zero. Hence, there exists some Brownian motion, denoted by $B$, s.t. $M_t=B_{\langle M\rangle_t}$ for all $t\ge 0$. Therefore, $X$ is equal in law to the following stochastic process :

$$Y_t=1+bt+W_{f(t)} \quad \mbox{with}\quad f(t):=\langle M\rangle_t=\int_0^t\frac{\sigma^2}{(1+m(s))^2}ds,\quad \forall t\ge 0.$$

Hence,

$$\mathbb P[\inf_{0\le s\le t}X_s>0]=\mathbb P[\inf_{0\le s\le t}Y_s>0]=\mathbb P\big[\inf_{0\le u\le f(t)}Y_{f^{-1}(u)}>0\big]=\mathbb P\big[\inf_{0\le u\le f(t)}\big(1+bf^{-1}(u)+W_u\big)>0\big].$$

Let $\lambda:=(f^{-1})'$ (more precisely $\lambda(u)=\big(1+m(f^{-1}(u))\big)^2\big/\sigma^2$) and define the change of measure by

$$\frac{d\mathbb Q}{d\mathbb P}:=\exp\left(-b\int_0^{f(t)}\lambda(u)dW_u-\frac{b^2}{2}\int_0^{f(t)}\lambda(u)^2du\right).$$

Then $(Z_u:=bf^{-1}(u)+W_u)_{u\ge 0}$ is a Brownian motion under $\mathbb Q$. Further,

$$\mathbb P\big[\inf_{0\le u\le f(t)}\big(1+bf^{-1}(u)+W_u\big)>0\big]=\mathbb E^{\mathbb Q}\left[\frac{d\mathbb P}{d\mathbb Q}{\bf 1}_{\{\inf_{0\le u\le f(t)}(1+Z_u)>0\}}\right]=\mathbb E^{\mathbb Q}\left[\exp\left(b\int_0^{f(t)}\lambda(u)dZ_u-\frac{b^2}{2}\int_0^{f(t)}\lambda(u)^2du\right){\bf 1}_{\{\inf_{0\le u\le f(t)}Z_u>-1\}}\right].~~~~ (\ast)$$

Here $(\ast)$ provides an upper bound of the probability. To compute the probability, it remains to find out the joint law of

$$\left(\int_0^t \lambda(s)dZ_s, \inf_{0\le s\le t}Z_s\right).$$

Is this known in the literature?