Specific impulse is usually defined as $I_{sp} = \frac{F_T}{\dot m ~ g_0} $
That's true only if you use standard metric units. With force expressed in pounds-force and mass flow rate expressed in pounds, one simply divides the force (in pounds-force) by the mass flow rate (in pounds/second) and voila! you have specific impulse in lbf·s/lb. For example, the first stage of the Saturn V produced 7,715,150 pounds-force of thrust at launch while consuming fuel at a rate of 29,157.58 pounds/second. Divide 7,715,150 by 29,157.58 and you get 264.6, the specific impulse at launch. Properly, this value of 264.6 is in units of lbf·s/lb. If you do the math, it is also numerically equal to the specific impulse in seconds.
Alternatively, one could convert that force to newtons and the mass to kilograms, yielding 34.3817 meganewtons of thrust and 13.22565 metric tons per second of fuel consumption. Now the division yields 2594.9 m/s. Divide by g0=9.80665 m/s2 and you get 264.6 seconds.
A second alternative is to convert that force in pounds-force to kilogram-force. This conversion yields a force of 3,499,530 kilograms-force. Now we're back to using the trick of simply dividing the force (in kg-f) by the mass flow rate (in kg/s): 3499530/13225.65 = 264.6.
Germany was the leading European developer of rocketry up until 1945. They used kilogram-force to express thrust rather than newtons. The Americans and Russians took over after that point. The Americans stuck to using customary units, pounds-force and pounds-mass. The Russians followed the German tradition of expressing thrust in kilograms-force. In both cases, simply dividing force by mass yields specific impulse in seconds.
In fact, despite being a banned unit, most European aerospace engineers tended to express thrust in kilograms-force rather than newtons up until the 1980s, and some still do. It is a very convenient unit for spacecraft and aircraft that operate near the Earth.
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I think you've misread the article. It says rocket engines can attain up to 70% $\eta_c$, which is only the cycle efficiency (how well it turns the energy of the fuel into mechanical energy). This is not the propulsive efficiency.
Unfortunately, for a rocket much of this mechanical energy is used to (wasted..) increase the KE of the exhaust rather than the rocket. As the article mentions, optimum efficiency is when the exhaust speed and rocket speed are matched. But this ends up being horrible for fuel consumption.
Being able to throw the mass of the earth or the atmosphere around makes regular propulsion much more efficient.
In one of your comments you linked to the question Velocity and kinetic energy, violating galilean relativity and said that the efficiency of a car drops with speed. I wouldn't agree with that statement. The question was specifically about interpreting energy in different frames.
If we stick to to just the frame where the ground is at rest (a very valid frame for travel on the earth), then the theoretical efficiency of your battery car approaches 1 as you eliminate drag. The energy of the battery can be given into KE of the vehicle almost entirely since the earth is so massive.