Ion thrusters are usually specified with two numbers: thrust and electrical power. As an example, let's take the NSTAR engine of DS1. According to Wikipedia it produced 92 mN thrust at a power of 2.3 kW.
Now we can apply Newtons well-known formula to determine the acceleration
$$F = m \cdot a \quad \rightarrow a = \frac{F}{m}$$
as well as the velocity change if the engine runs for a time T:
$$\Delta v = a \cdot T = T \cdot \frac{F}{m}$$
Let's call the 'efficiency' of the engine $\epsilon = \frac{F}{P}$, i.e. the force generated from a given amount of power. Please note that this is not a fixed number, but varies - doubling the power usually does not double the force.
$$\Delta v = \epsilon \cdot P \cdot T \cdot \frac{1}{m}$$
or, in terms of energy expended:
$$E = P \cdot T = \frac{m}{\epsilon}\cdot \Delta v$$
As an example, to accelerate DS 1 (m = 500 kg) by $\Delta v$ = 100 m/s, we need:
$$E = \frac{m}{\epsilon}\Delta v = m \cdot \frac{P}{F} \cdot \Delta v$$
$$\quad = 500 \rm{kg} \cdot \frac{2.3 \rm{kW}}{92\rm{mN}}\cdot 100\frac{\rm m}{\rm s} = 1250 MWs = 347 kWh $$
As you can see, we are assuming constant vehicle mass and are not employing the rocket equation. When using ion thrusters, the fuel consumption is quite low for smaller adjustments - in our example about 0.3% of the total vehicle mass. For larger $\Delta v$ we have to come back to the conventional rocket equation - or do a break down manually by dividing the change into several smaller changes.
The diagram you cite is unfortunately not applicable for low-thrust ion propulsion. It is only valid for instantaneous propulsive manoveurs, but not for extended burn times as needed here. In general, the required $\Delta v$ is larger the lower the thrust is, but the precise numbers have to be calculated using all the details of the planned journey and can't be predicted from a simple chart like yours.
