$\newcommand\ep\varepsilon$The case of interest when $b(t,x)=b(t)$ depends only on $t$ is comparatively simple.

Indeed, let
$$g(t):=\sqrt{(2t+1/2)\ln\ln(3+t)}$$
for real $t\ge0$.
By the law of the iterated logarithm,
$$\sup_{s\in[t,\infty)}\frac{W_s}{g(s)}\to1$$
as $t\to\infty$ almost surely and hence in probability.

So, for each real $\ep>0$ there is some real $t=t_\ep>0$ such that
\begin{equation*}
\tfrac\ep2\,g(t)>(1+\ep)g(0) \tag{1}\label{1}
\end{equation*}
and
\begin{equation*}
P(B)>0, \tag{2}\label{2}
\end{equation*}
where
\begin{equation*}
B:=\{W_s<(1+\ep/2)g(s)\ \forall s\in[t,\infty)\}.
\end{equation*}
Let
\begin{equation*}
A:=\{W_s<(1+\ep)g(s)\ \forall s\in[0,t]\}.
\end{equation*}
Note that the function $g$ is concave. So, for all $s\in[0,t]$ we have $g(s)\ge g(0)+\frac st\,(g(t)-g(0))$ and hence for all real $u<(1+\ep/2)g(t)$
\begin{equation*}
\begin{aligned}
&(1+\ep)g(s)-\tfrac st\,u \\
&\ge(1+\ep)g(0)+\tfrac st\,[(1+\ep)(g(t)-g(0))-u] \\
&\ge(1+\ep)g(0)+\tfrac st\,[(1+\ep)(g(t)-g(0))-(1+\ep/2)g(t)] \\
&=(1+\ep)g(0)+\tfrac st\,[\tfrac\ep2\,g(t)-(1+\ep)g(0)] \\
&\ge(1+\ep)g(0)>g(0),
\end{aligned}
\tag{3}\label{3}
\end{equation*}
in view of \eqref{1}.

Note also that the Brownian bridge $W^{(t)}_\cdot$ defined by the formula $W^{(t)}_s:=W_s-\tfrac st\,W_t$ for $s\in[0,t]$ is independent of $W_t$.
Recalling also the symmetry of $W_\cdot$ and its Markov property, as well as \eqref{3}, we get
\begin{equation*}
\begin{aligned}
&P(W_s>-(1+\ep)g(s)\ \forall s\in[0,\infty)) \\
&=P(W_s<(1+\ep)g(s)\ \forall s\in[0,\infty)) \\
&=P(A\cap B) \\
&=P(W_s<(1+\ep)g(s)\ \forall s\in[0,t],W_t<(1+\ep)g(t),B) \\
&=\int_{-\infty}^{(1+\ep)g(t)} P(W_t\in du,B)
P(W_s-\tfrac st\,W_t<(1+\ep)g(s)-\tfrac st\,u\ \forall s\in[0,t]) \\
&=\int_{-\infty}^{(1+\ep)g(t)} P(W_t\in du,B)
P(W^{(t)}_s<(1+\ep)g(s)-\tfrac st\,u\ \forall s\in[0,t]) \\
&\ge\int_{-\infty}^{(1+\ep)g(t)} P(W_t\in du,B)
P(W^{(t)}_s<g(0)\ \forall s\in[0,t]) \\
&=P(B)
P(W^{(t)}_s<g(0)\ \forall s\in[0,t])>0,
\end{aligned}
\tag{4}\label{4}
\end{equation*}
by \eqref{2} and because $g(0)>0$.

Note that $X_s=1+B(s)+W_s$. So,

\begin{equation*}
\begin{aligned}
&P(X_s>1+B(s)-(1+\ep)g(s)\ \forall s\in[0,\infty)) \\
&=P(W_s>-(1+\ep)g(s)\ \forall s\in[0,\infty))>0,
\end{aligned}
\end{equation*}
by \eqref{4}.
Thus, the condition that
\begin{equation*}
B(s)>(1+\ep)g(s)-1 \tag{5}\label{5}
\end{equation*}
for some real $\ep>0$ and all $s\in[0,\infty)$ is sufficient for
\begin{equation*}
P(X_s>0\ \forall s\in[0,\infty))>0.
\end{equation*}
Note finally that for any $\ep\in(0,\frac1{g(0)}-1)=(0,3.61\ldots)$ there is a positive continuous function $b$ such that the function $B$ given by the formula $B(t)=\int_0^t b(s)\,ds$ for all real $t\ge0$ satisfies condition \eqref{5} (however, of course, $b$ does not have to be everywhere positive or continuous in order for $B$ to satisfy condition \eqref{5}).