Electronic – How to handle delta function after finding the impulse response

\$h(t) = \int_{-\infty}^t \! \delta(T-2)e^{-(t-T)} \, \mathrm{d}T = e^{-(t-2)}u(t-2)\$

The delta "function" is zero except where the argument is zero, i.e., when T=2, where it has an area of 1.

So, if \$ t < 2\$, the integral is zero.

If \$ t \ge 2\$, the integral equals the area of the delta function multiplied by the value of the exponential when T = 2.

Assuming that with "associate" you mean "put in series" that is you hook S1 output to S2 input then you are not doing anything wrong. Problem is that sometimes (plenty of times) solving a convolution integral is not a piece of cake thus you prefere the Laplace transform.

Let's start by computing \$G_1(s)\$ and \$G_2(s)\$:

\$g_2(t)=e^{-\alpha t}u(t)\$, but \$U(s)=\mathcal{L}\{u(t)\}=\frac{1}{s}\$ and you can use the frequency shifting property that goes: $$ e^{at}f(t) \leftrightarrow F(s-a) $$ finally: $$ G_2(s)= \frac{1}{s+\alpha} $$

That's called exponential decay and is usually provided in Laplace transforms tables. Of course \$G_1(s)=G_2(s)|_{\alpha=1}\$ so that: $$ G_1(s) = \frac{1}{s+1} $$ Now we're searching for $$G(s)\triangleq G_1(s)G_2(s)=\frac{1}{s+1}\frac{1}{s+\alpha}=\frac{A}{s+1}+\frac{B}{s+\alpha}=\frac{s(A+B)+A\alpha+B}{(s+1)(s+\alpha)}$$ From the last step you can find out \$A=-B=\frac{1}{\alpha-1}\$, now we're at: $$ G(s) = \frac{1}{\alpha-1}\left(\frac{1}{s+1}-\frac{1}{s+\alpha}\right) $$

The anti transform is quite simple now since both terms in parenthesis are known transforms:

$$ g(t) = u(t)\frac{1}{\alpha-1}\left(e^{-t}-e^{-\alpha t}\right) $$

Could you get this result from your integral? Probably. Did you? Nope, and that's because integrals are so boring. Plus your book/teacher probably wanted you to use the whole A and B thing, also known as partial fraction decomposition.

The Laplace transform is a great weapon, learn how to use it and you will be rewarded.