Consider the folloing ODE system: x ′ ( t ) = A ( t )...

We assume that $b,A$ are continuous over $\mathbb{R}$ and that (1) $t\in \mathbb{R}\to |{\int }_{t_0}^{t}b|$ is bounded by $K$ and (2) $t\in \mathbb{R}\to |{\int }_{t_0}^t||A|||$ is bounded by $L$.
Then $x(t)={\int }_{t_0}^{t}Ax+{\int }_{t_0}^{t}b+x_0$ and $||x(t)||\le {\int }_{t_0}^t||A||||x||+K+||x_0||$; according to Gronwall, $||x(t)||\le (K+||x_0||)\exp ({\int }_{t_0}^t||A||)\le (K+||x_0||)\exp (L)$. Thus the solution $x(t)$ is bounded over any segment and, consequently, the maximal solution is defined over whole $\mathbb{R}$.
EDIT. We can do better (although I feel that my post does not interest the OP). The conditions (1),(2) are so that the solution exists and is bounded over $\mathbb{R}$. If you want only the existence, then (1),(2) are useless.
Proof. Let $I$ be a segment, $K_I=\underset{t\in I}{sup}|{\int }_{t_0}^{t}b|,L_I=\underset{t\in I}{sup}|{\int }_{t_0}^t||A|||$. Then $\underset{t\in I}{sup}||x(t)||\le (K_I+||x_0||)\exp (L_I)$ and we are done.