There is a geometric way of seeing that inversion works properly, which is based on considering not $1/z$ but $\bar{1/z}$. Then your $(x,y)$ gets mapped to $(\frac x{x^2+y^2},\frac y{x^2+y^2})$, which can be interpreted geometrically as follows. Let $O$ be the origin, and let $P$ be a point. Then $P$ gets sent to the unique point P' on the ray $\vec{OP}$ such that OP\cdot OP'=1 (where $OP$ is the length of the segment joining $O$ and $P$). If $R$ is any point on the unit circle between $P$ and P', then we can rewrite the equation as \frac{OP}{OR}=\frac{OR}{OP'}. Hence, triangles $OPR$ and ORP' will be similar. You can then prove that triangles $OPQ$ and OQ'P' are similar for any points $P$ and $Q$. (multiply to get $\frac{OP}{OR}\frac{OR}{OQ}=\frac{OQ'}{OR}\frac{OR}{OP'}$).
Once you have that it is easy to show that if a line $\ell$ does not go through the origin, it gets mapped to a circle that goes through the origin as follows. Suppose that $P$ is on $\ell$ such that $\vec{OP}$ is perpendicular to $\ell$. Then for any other point $Q$ on $\ell$ we have that triangle $OPQ$ has right angle $OPQ$, and hence triangle OQ'P' has right angle OQ'P' for all points Q'. But that implies that the points Q' trace out a circle with diameter OP'. Hence, $\vec{OP}$, which was perpendicular to $\ell$ is perpendicular to the circle that $\ell$ got sent to. This shows full conformality since if you have two lines $\ell_1$ and $\ell_2$, you have three cases.
- $\ell_1$ and $\ell_2$ go through the origin -- they are fixed by inversion.
- $\ell_1$ goes thorugh the origin, but $\ell_2$ does not. Then \ell_2' which is the perpenciular line to $\ell_2$ though the origin is fixed along with $\ell_1$ and perpendicular to the image-circle of $\ell_2$, hence the angle between $\ell_1$ and the circle equals the angle between $\ell_1$ and $\ell_2$.
- Both don't go through the origin, in which case take both of their perpendiculars $\ell_1$ and $\ell_2$.
There is also the analytic way of doing this.
If we consider $\mathbb C$ as the real plane $\mathbb R^2$, then what you're trying to show is that certain functions $f\colon \mathbb R^2\to\mathbb R^2$ are angle-preserving.
To be angle preserving at a particular point means that the derivative $df(z_0)$, which is a $2\times 2$ real matrix, preserves the angles between vectors. To see this, take two (smooth) curves $c_1\colon\mathbb R\to\mathbb R^2$ and $c_2\colon\mathbb R\to\mathbb R^2$. The angle between them at point $z_0=c(t_1)=c(t_2)$ is, as you said, the angle between the tangent lines, which is of course the angle between the tangent vectors $dc_1(t_1)$ and $dc_2(t_2)$ -- let's call them $v_1$ and $v_2$.
The function $f$ will map the two curves to the curves $f\circ c_1$ and $f\circ c_2$, whose tangent vectors will be $df(z_0)v_1$ and $df(z_0)v_2$. Thus, we want $df(z_0)$ to be a matrix $M$ that preserves the angles between vectors.
Fortunately, you're not asking how to classify all conformal mappings, so we just need to check that the derivatives of $z\to az+b$, $z\to 1/z$ and $z\to\bar z$ preserve angles at every $z_0$.
It turns out that the Cauchy-Riemann equations tell you that complex-differentiable functions (such as $z\to az+b$ and $z\to 1/z$) have derivates that look like $M=\left[\begin{matrix}a&b\\-b&a\\\end{matrix}\right]$ which since $\det M=a^2+b^2$ look exactly like $a^2+b^2\left[\begin{matrix}\cos(\theta)&\sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{matrix}\right]$ for some angle $\theta$, which is exactly dilation by a factor of $a^2+b^2$ and rotation by angle $\theta$, which evidently preserve angles between vectors (for a quick proof, the rotation matrix is orthonormal and so actually preserves dot products, while dilation merely scales dot products by some constant).
To get the action of $\bar z$, note that it sends $x+iy$ to $x-iy$ and so its derivative is the matrix $\left[\begin{matrix}1&0\\0&-1\end{matrix}\right]$. This matrix does not preserve the dot product -- a short computation shows that it reverses the sign of the dot product. But that is ok since reversing the sign of the dot product is a reversal of orientation, i.e. the angle changes from $\theta$ to $180^o-\theta$, which for the purposes of being conformal is the same thing.