If $b_2$ divides $b_1$ then an obvious necessary and sufficient condition is that $a_1\not\equiv a_2\ \mathrm{mod}\ b_2$.

If $b_2$ does not divide $b_1$ then $\gcd(b_1,b_2)$ is a proper divisor of $b_2$, so there are at least two residue classes $y\ \mathrm{mod}\ b_2$ such that $y\equiv a_1\ \mathrm{mod}\ \gcd(b_1,b_2)$. Pick one that satisfies $y\not\equiv a_2\ \mathrm{mod}\ b_2$, then by the criterion you quote there is an $x$ satisfying $x\equiv a_1\ \mathrm{mod}\ b_1$ and $x\equiv y\ \mathrm{mod}\ b_2$. This $x$ solves the second system.

To summarize, the second system has a solution unless $b_2\mid b_1$ and $a_1\equiv a_2\ \mathrm{mod}\ b_2$.