How does an ideal op amp amplify a voltage input when the voltage difference is 0?
A transfer function of closed loop system is:

\$H(s) = \dfrac{F(s)}{1+F(s)}\$







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A transfer function of closed loop system  with a feedback is:

\$H(s) = \dfrac{F(s)}{1+F(s)\cdot G(s)}\$







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Now let's suppose you have a opamp buffer - the output is feedback to inverting input, as case 1 in this answer. F(s) = G (Gain), you can notice that output is never equal to the voltage on non-inverting input, but very close. 







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The error is :

\$\varepsilon = \dfrac{1}{1+G}\$

So, the voltages on both inputs are never equal, but very close, and higher gain makes them closer. 
Since the opamp's gain is insane high, for the sake of simplicity we can say that voltage of inverting input is equal to the voltage of the non-inverting input. If you ground the non-inverting input, then you get a virtual ground on the inverting input, but it is never zero voltage for real. And as you already discovered it also can't be zero, else no signal would come out.