@Old John's traditional geometry answer is obviously all right and, well... geometric, but, if one needs to be reassured, the Cartesian brute force approach works as well.
To see this, assume without loss of generality that $a$ is a real number, then $z=x+\mathrm iy$ yields $|z\pm a|=\sqrt{x^2\pm2ax+a^2+y^2}$. Squaring this, one sees that the desired set has equation $ x^2\pm2ax+a^2+y^2=|z\pm a|^2\leqslant(2b-|z\mp a|)^2=4b^2+x^2\mp2ax+a^2+y^2-4b|z\mp a|, $ that is, $ b^2\mp ax\geqslant b|z\mp a|, $ or, squaring again, $ b^4\mp2b^2ax+a^2x^2\geqslant b^2(x^2\mp2ax+a^2+y^2), $ that is, $ (b^2-a^2)x^2+b^2y^2\leqslant b^2(b^2-a^2), $ which describes the interior of an ellipse with the coordinate axes as principal axes and the origin as center. Equivalently, $ \frac{x^2}{b^2}+\frac{y^2}{c^2}\leqslant1,\qquad\text{with}\quad c^2=b^2-a^2. $