y(t) is not

\(\displaystyle y = A e^t + B e^{-t}\)

It is

\(\displaystyle y = (A e^t + B e^{-t})u(t)\)

where u(t) is the unit step function. This is true, because a causal system excited at t = 0 by an impulse cannot have a response before t = 0.

Now, on to your question. We will need to know also the form of y'(t):

\(\displaystyle y' = (A e^t + B e^{-t})\delta(t) + (Ae^t-Be^{-t})u(t)\)

We then integrate both sides of the equation from 0- to 0+:

\(\displaystyle \int_{0^-}^{0^+}y''(t)\, dt-\int_{0^-}^{0^+}y(t)\, dt=\int_{0^-}^{0^+}\delta(t)\, dt\)

which becomes:

\(\displaystyle y'(0+) - y'(0-) - \int_{0^-}^{0^+}y(t)\, dt=u(0+) - u(0-)\)

Now, it is obvious from looking at the form of y(t) that it has no time to develop area nor does it have any higher order singularity functions. Therefore, an integral from 0- to 0+ yields zero. The unit step is 1 at 0+ yet 0 at 0-:

\(\displaystyle y'(0+) - y'(0-)=1\)

Now, the impulse is zero at both 0- and 0+, and the u(t) is 1 at 0+ and 0 at 0-. Therefore, y'(0-) = 0. Further, y'(0+) = ae^0+ - be^0+ = a - b. So we can finally write:

\(\displaystyle A - B = 1\)

To find the next equation, you need to integrate twice. We can do that one a little faster since we've gone through some of the reasoning:

integrating twice yields y(0+) - y(0-) + 0 = 0

then A + B = 0

edit: Also, an impulse is another name for the dirac delta. impulse integrated = u(t). u(t) integrated = ramp(t) = t u(t). That's how I said the double integration equals zero. It was a double integral of delta(t) = ramp(t), and ramp(0+) = ramp(0-) = 0.