How to measure energy of 120/250V DC, 100 ampere system
As Jasen suggests - unless you wish to build an energy meter for 
the experience then you can do it at less cost and much more easily using a premade Chinese (usually) unit.
The example he gives looks suitable for your purpose and is very well priced.

It is EXTREMELY IMPORTANT to TAKE EXCEPTIONAL CARE when dealing with DC at the voltages and current you mention.  


120 VDC can kill you instantly - and 250 VDC even faster  :-) :-(.
And 100A DC can kill you and burn down youR installation / premises / home as a bonus. 


Whereas 100A AC is just "very dangerous" but not TOO hard to deal with, 100A DC at 100V+ is extremely hard to turn off. Switches and breakers MUST be rated to handle the voltages and current concerned. Fuses must be appropriately rated HRC ones suited to the application.  While an AC arc will (usually) self extinguish, a DC one usually won't without specific design steps to ensure that this happens.

USE "SERIOUS" ISOLATION

When using power meters with say 48V or higher 10A or higher DC,
 I VERY STRONGLY recommend that you use an isolated link - such as a radio link (such as is used in the mentioned Chinese unit) or optocouplers rated appropriately or similar.

______________________________________

NEVER DEPEND ON 'ISOLATION' OFFERED BY CURRENT CARRYING SENSORS

Depending on the isolation (allegedly) provided by Hall effect magnetic field sensors with high voltage isolation is not at all wise. Hall sensors usually have an integral "zero Ohm," high current link in the sensor to allow the DC current path location to be well defined relative to the Hall sensor cell. Under fault conditions, if currents well in excess of rated current flows the link within the sensor may melt and destroy the sensor - either causing a fire or just a molten pool of sensor stuff, or both. 
This has the potential to allow the measured circuit and the measurement circuit to "become one" - with potentially disastrous and costly and even possibly fatal consequences.
Ask me how I know :-) :-(.
I had such an event occur - fortunately with not overly expensive consequences - but it could have easily enough destroyed a large amount of low voltage equipment. 

Sensors which do not carry high currents and which BY DESIGN cannot be destroyed in a manner that breaches the isolation barrier are 'safe' to use. (for most values of safe). 



The vital qualification is "CURRENT CARRYING" sensors where the current path is part of the sensor itself and in closely connected mechanically and thermally.  This is an example of the type of sensor I mean.
This is an Allegro ACS756SCB-100B-PFF-T Hall effect current sensor, rated at 100 A.
The device will withstand 3 kVAC applied between the current path and the electronics - but only as long as the whole device has been melted into a pile of smoking slag by a high primary current caused by a short circuit. Ask me how I know :-) :-(.  

The current to be measured (&lt;= 100A) travels via the large "legs" and through an internal conductor of similar size. This is "safe" at 100A but not necessarily so at 500A or more fault currents. Any system using batteries and capable of carrying 100A is liable to have potential fault currents many times this value. 



This usually relates to magnetic field sensing devices which provide a current path within the sensor so that the location of the eg Hall cell is fixed relative to the conductor and the conductor shape is thus standardised.
The majority of Hall sensors for current measurement tend to fall in this category. 

At their rated current (1A 10A 100A ....) the internal path resistance is low enough that heating is not a problem. But if a say 50A sensor has a fault current of 500A (say a downstream short in a battery circuit then fault current is 500/50 = 10x rated and heating from I^R is 100x rated and ... Wow!!!  

Say a 50A sensor has a 1 mOhm resistance.
At 50A, Vdrop = 0.05V and heating at 50A = 2.5W (!).
At 500 A heating is 250W.
It won't last long :-).   

Even at 100A W = 10W. At 200A it's 40W.   

I Was working with a 40 Ah x 24V LiFePO4 battery in a development project.
I don't know what the short current was that slagged the sensor and "ohmically united" the high and low voltage circuits, BUT I was "just lucky" that the 24V did not get very far into my low voltage control circuitry.   

Isolation COULD be in the LV interface so it keeps any HV out of associated circuitry. Easy enough, but essential.