Kepler's equation: $M=E-\varepsilon\cdot \sin( E)$, you need to solve for $E$.
The inverse problem may be solved (according to wikipedia) as follows:
$E = \begin{cases} \displaystyle \sum_{n=1}^{\infty} {\frac{M^{\frac{n}{3}}}{n!}} \lim_{\theta \to 0} \left( \frac{\mathrm{d}^{\,n-1}}{\mathrm{d}\theta^{\,n-1}} \left( \frac{\theta}{ \sqrt[3]{\theta - \sin(\theta)} } ^n \right) \right) , & \epsilon = 1 \\ \displaystyle \sum_{n=1}^{\infty} { \frac{ M^n }{ n! } } \lim_{\theta \to 0} \left( \frac{\mathrm{d}^{\,n-1}}{\mathrm{d}\theta^{\,n-1}} \left( \frac{ \theta }{ \theta - \epsilon \cdot \sin(\theta)} ^n \right) \right) , & \epsilon \ne 1 \end{cases}$
which evaluates to:
$E = \begin{cases} \displaystyle M + \frac{1}{60} M^3 + \frac{1}{1400}M^5 + \cdots \ | \ x = ( 6 M )^\frac{1}{3} , & \epsilon = 1 \\ \\ \displaystyle \frac{1}{1-\epsilon} M - \frac{\epsilon}{( 1-\epsilon)^4 } \frac{M^3}{3!} + \frac{(9 \epsilon^2 + \epsilon)}{(1-\epsilon)^7 } \frac{M^5}{5!} + \cdots , & \epsilon \ne 1 \end{cases} $
The above is copied straight out of wikipedia. It's kind of ugly. Plus, the first equation is written in "x" instead of "M" and you can't be sure it's correct.
So instead, what I'd try is to take the first terms from the above, i.e. put $E_0 = M$ or $M/(1-\epsilon)$, and use Newton's method to iterate. That is, iterate with:
$E_{n+1} = E_n - \frac{M - E +\epsilon \sin E}{-1+\epsilon\cos(E)}$.
To test the method, run a bunch of random data through it, and make sure you test the fence posts (boundary conditions).
For the case of hyperbolic orbit, you use the hyperbolic trigonometric function. So your equation for Hyperbolic Mean anomaly becomes:
$$M_h = e\sinh F - F$$
where $F$ is your hyperbolic eccentric anomaly, which is analogous to eccentric anomaly for ellipse and $M_h$ is still given by the same expression. The angle measurement for hyperbolic trigonometric identities are not the same but analogous, so using the center of hyperbola which exists between the real hyperbola and it's image, you get the sinh function value. Assuming the object to be at distance $r$ from the primary focus and center of hyperbola being the reference origin $\sinh F$ will be given by:
$$\sinh F = \dfrac{y}{b}$$
where $y$ is the perpendicular distance from the focal axis of the object on trajectory, and $b$ is the semi-minor axis of the conic.
As for the semi-major axis it can be obtained by orbit equation at periapse for hyperbola which is $a = \dfrac{h^2}{\mu} \dfrac{1}{e^2-1}$, which is positive.
For parabolic orbit you do not define Mean Anomaly, because of rightly mentioned absence of geometric center, instead use Baker's equation $2\sqrt{\dfrac{\mu}{p^3}}\left(t-\tau\right) = \tan\dfrac{\theta}{2} +\dfrac{1}{3}\tan^3\dfrac{\theta}{2}$
where $\tau$ is your periapse passage time and $t$ is the elapsed time since periapse, and $p$ is the semi-latus rectum which is well defined for a parabola. There are solutions available to this equation that gives the position of the object at any given time given some input parameters.
Hope this helps.
Best Answer
As far as I can tell the true anomaly is the same type of angle used in the standard solution of the Kepler problem since there we assume the sun is at a foci. When solving the equations of motion for a Keplerian orbit we obtain $r\left(\theta\right)=\frac{a\left(1-e^{2}\right)}{1\pm e\cos\theta}$ (- if $r\left(0\right)$ is through the origin and + if it is away from the origin) and we can express the velocity as $v=\sqrt{\mu\left(\frac{2}{r}-\frac{1}{a}\right)}$ where $\mu$ is the standard gravitational parameter which in the two body case is just $G\left(m_{1}+m_{2}\right)$.
The derivation of these formulae can be found in many mechanics textbooks. For example, Taylor - Classical Mechanics - chapter 8.