= Solution
The image is a <ring torus>: a circle of radius $b$ is rotated about the vertical axis at distance $a$ from that axis. Put $r=\sqrt{x^2+y^2}$. Since $a>b$, every image point has $r>0$, and the image is precisely
$$
T=\{(x,y,z):(r-a)^2+z^2=b^2\}.
$$
Conversely, on this <level set> the pairs $((r-a)/b,z/b)$ and $(x/r,y/r)$ are unit vectors, so choosing their angles gives a preimage.
The specified angular rectangle removes two closed circles from the <torus>. More explicitly its image is
$$
U=T\setminus(C_u\cup C_v),\qquad
C_u=\{z=0,\ r=a+b\}\cap T,\quad
C_v=\{y=0,\ x>0\}\cap T.
$$
Here $C_u$ is the outer equatorial circle and $C_v$ is a meridian. The given openness property makes $U$ open in $T$. On $U$, the angles of $(r-a)+iz$ and $x+iy$ are uniquely chosen in $(0,2\pi)$; their continuous argument branches give the inverse $(x,y,z)\mapsto(u,v)$. Thus the restriction is a <homeomorphism> onto $U$, and the inverse is locally <smooth>.
For the <immersion> condition, the two derivatives have <inner products>
$$
\sigma_u\cdot\sigma_u=b^2,\qquad
\sigma_v\cdot\sigma_v=(a+b\cos u)^2,\qquad
\sigma_u\cdot\sigma_v=0.
$$
They are <linearly independent> everywhere, since $a+b\cos u\ge a-b>0$. Hence this is a smooth <embedded surface parametrization>.
The angular cuts do not represent singularities of the <embedded torus of revolution>. To give a direct proof covering all points, define $F=(\sqrt{x^2+y^2}-a)^2+z^2-b^2$ on the open region $r>0$. On $T$,
$$
\nabla F=2\left((r-a)\frac{x}{r},(r-a)\frac{y}{r},z\right),\qquad
|\nabla F|=2b>0.
$$
The <implicit function theorem> therefore describes $T$ near every point as a <smooth> graph of two coordinates. \b[The whole torus is a smooth embedded surface.]
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