Hi Anweshi,
Since Emerton answered your third grey-boxed question very nicely, let me try at the first two. Suppose $L(s,f)$ is one of the L-functions that you listed (including the first two, which we might as well call L-functions too). (For simplicity we always normalize so the functional equation is induced by $s\to 1-s$.) This guy has an expansion $L(s,f)=\sum_{n}a_f(n)n^{-s}$ as a Dirichlet series, and the most general prime number theorem reads
$\sum_{p\leq X}a_f(p)=r_f \mathrm{Li}(x)+O(x \exp(-(\log{x})^{\frac{1}{2}-\varepsilon})$.
Here $\mathrm{Li}(x)$ is the logarithmic integral, $r_f$ is the order of the pole of $L(s,f)$ at the point $s=1$, and the implied constant depends on $f$ and $\varepsilon$.
Let's unwind this for your examples.
1) The Riemann zeta function has a simple pole at $s=1$ and $a_f(p)=1$ for all $p$, so this is the classical prime number theorem.
2) The Dedekind zeta function (say of a degree d extension $K/\mathbb{Q}$) is a little different. It also has a simple pole at $s=1$, but the coefficients are determined by the rule: $a(p)=d$ if $p$ splits completely in $\mathcal{O}_K$, and $a(p)=0$ otherwise. Hence the prime number theorem in this case reads
$|p\leq X \; \mathrm{with}\;p\;\mathrm{totally\;split\;in}\;\mathcal{O}_K|=d^{-1}\mathrm{Li}(x)+O(x \exp(-(\log{x})^{\frac{1}{2}-\varepsilon})$.
This already has very interesting applications: the fact that the proportion of primes splitting totally is $1/d$ was very important in the first proofs of the main general results of class field theory.
3) If $\rho:\mathcal{G}_{\mathbb{Q}}\to \mathrm{GL}_n(\mathbb{C})$ is an Artin representation then $a(p)=\mathrm{tr}\rho(\mathrm{Fr}_p)$. If $\rho$ does not contain the trivial representation, then $L(s,\rho)$ has no pole in neighborhood of the line $\mathrm{Re}(s)\geq 1$, so we get
$\sum_{p\leq X}\mathrm{tr}\rho(\mathrm{Fr}_p)=O(x \exp(-(\log{x})^{\frac{1}{2}-\varepsilon})$.
The absence of a pole is not a problem: it just means there's no main term! In this particular case, you could interpret the above equation as saying that "$\mathrm{tr}\rho(\mathrm{Fr}_p)$ has mean value zero.
4) For an elliptic curve, the same phenomenon occurs. Here again there is no pole, and $a(p)=\frac{p+1-|E(\mathbb{F}_p)|}{\sqrt{p}}$. By a theorem of Hasse these numbers satisfy $|a(p)|\leq 2$, so you could think of them as the (scaled) deviation of $|E(\mathbb{F}_p)|$ from its
"expected value" of $p+1$. In this case the prime number theorem reads
$\sum_{p\leq X}a(p)=O(x \exp(-(\log{x})^{\frac{1}{2}-\varepsilon})$
so you could say that "the average deviation of $|E(\mathbb{F}_p)|$ from $p+1$ is zero."
Now, how do you prove generalizations of the prime number theorem? There are two main steps in this, one of which is easily lifted from the case of the Riemann zeta function.
Prove that the prime number theorem for $L(s,f)$ is a consequence of the nonvanishing of $L(s,f)$ in a region of the form $s=\sigma+it,\;\sigma \geq 1-\psi(t)$ with $\psi(t)$ positive and tending to zero as $t\to \infty$. So this is some region which is a very slight widening of $\mathrm{Re}(s)>1$. The proof of this step is essentially contour integration and goes exactly as in the case of the $\zeta$-function.
Actually produce a zero-free region of the type I just described. The key to this is the existence of an auxiliary L-function (or product thereof) which has positive coefficients in its Dirichlet series. In the case of the Riemann zeta function, Hadamard worked with the auxiliary function $ A(s)=\zeta(s)^3\zeta(s+it)^2 \zeta(s-it)^2 \zeta(s+2it) \zeta(s-2it)$. Note the pole of order $3$ at $s=1$; on the other hand, if $\zeta(\sigma+it)$ vanished then $A(s)$ would vanish at $s=\sigma$ to order $4$. The inequality $3<4$ of order-of-polarity/nearby-order-of-vanishing leads via some analysis to the absence of any zero in the range $s=\sigma+it,\;\sigma \geq 1-\frac{c}{\log(|t|+3)}.$ In the general case the construction of the relevant auxiliary functions is more complicated. For the case of an Artin representation, for example, you can take $B(s)=\zeta(s)^3 L(s+it,\rho)^2 L(s-it,\widetilde{\rho})^2 L(s,\rho \otimes \widetilde {\rho})^2 L(s+2it,\rho \times \rho) L(s-2it,\widetilde{\rho} \times \widetilde{\rho})$. The general key is the Rankin-Selberg L-functions, or more complicated L-functions whose analytic properties can be controlled by known instances of Langlands functoriality.
If you'd like to see everything I just said carried out elegantly and in crystalline detail, I can do no better than to recommend Chapter 5 of Iwaniec and Kowalski's book "Analytic Number Theory."
Best Answer
This is an elaboration on my comment below John's answer. Its goal is to give a very brief summary of the history underlying the question; I hope that it is more or less correct.
I think that it is fair to say that from the beginning it was understood that non-vanishing on the line $\Re(s) = 1$ was the main requirement for proving the prime number theorem.
However, in Hadamard and de la Vallee Poussin's approaches, this non-vanishing was fed into an explicit formula (which gives a Fourier-type expansion of the prime counting function, or some variant), which was in turn obtained from the $\zeta$-function by various Mellin transform games. In particular, if I understand correctly, establishing the explicit formula involves pushing the contour of integration to the left of $s = 1$ (since it involves a sum over the zeroes of the zeta function), and so doesn't just involve analysis in the region $\Re(s) \geq 1$.
As noted in anon's answer, Landau formulated an approach to PNT via a Tauberian theorem involving just analysis on the region $\Re(s) \geq 1,$ but as well as the crucial condition of there being no zeroes on the line $\Re(s) = 1$, there was a growth condition.
In his paper, Wiener discusses earlier Tauberian approaches, including Landau's, and then refers to Ikehara's theorem (proved in Ikehara's thesis, I believe, under Wiener's supervision) as being the "true theorem" (I'm fairly confident that I'm remembering his language correctly here), i.e. the one with the correct condition, those conditions being simply that there are no zeroes on the line $\Re(s) = 1$.