It's the simplest one of them all, once you see it.
The Dirac delta is really what is called a linear functional. It is very straightforward. Essentially, it assigns the value of a function via the standard scalar product for functions. In other words, for a function f of a real argument, and \mathbb{R}\ni a \geq 0,
\delta_a(f) = \langle \delta(x-a), \ f(x) \rangle = \int_{0}^\infty \delta(x-a)f(x) \ dx = f(a).
This may all look daunting, but it's about the simplest thing you'll ever learn once you understand it.
Let's look at this again:
\int_{0}^{\infty} \delta(x-a) f(x) \ dx = f(a)
and try to apply it to finding Laplace transforms. We want the Laplace transform of \delta (t-a), a \geq 0 (which is really a misnomer, because \delta isn't really a function at all - you actually need to define a completely new transformation to deal with generalized functions, and it has different properties! I won't worry about that in this post, though).
From the definition of the Laplace transform (as you already have it), this is
{\cal L}\{ \delta(t-a) \} = \int_{0}^\infty \delta(t-a)e^{-st} \ dt.
But look at what I wrote above! From the definition of the delta disribution, this integral is trivial. Let f(t) = e^{-st}, so our integral is just
\int_{0}^\infty \delta(t-a)f(t) \ dt
which from the definition of the delta is just the
value of f(x) when x=a, or in other words, it's just f(a). In this case,
f(a) = e^{-as},
so we just get
{\cal L}\{\delta(t-a)\} = e^{-as}.
Now, of course, the question is how to find things like
{\cal L}\{ \delta(t-a) f(t) \},
but I'm not going to tell you how to do that (at least not yet). I want you to look at what I've posted above (specifically the identities concerning the delta distribution itself), and see if you can figure it out on your own. Post your work here and we'll comment!
