Consider the following simplified example of the same phenomenon, which many students find clarifying.
Let $f(n)=1$, if there are $n$ consecutive $7$s in the decimal expansion of $\pi$, and otherwise $f(n)=0$. Is this function computable?
A naive attempt to compute $f(n)$ would simply proceed to search $\pi$ for $n$ consecutive $7$s. If found, the algorithm outputs $1$, but otherwise....and then the naive algorithm doesn't seem to know when to output $0$, and so students sometimes expect that $f$ is not computable.
But actually, $f$ is a computable function. If it happens that there are arbitrarily long sequences of $7$s in the decimal expansion of $\pi$, an open question, then $f$ is the constant $1$ function, which is certainly computable. Otherwise, there is some longest sequence of $7$s in $\pi$, having length $N$, and so $f$ is the function that is $1$ up to $N$ and then $0$ above $N$. And this function also is computable, for any particular $N$.
So the situation is that we have proved that $f$ is computable by exhibiting several algorithms, and proving that $f$ is definitely computed by one of them, but we don't know which one. (In fact, $f$ is linear time computable.) So we have proved that $f$ is a computable function, but by a pure existence proof that merely shows there is an algorithm computing $f$, without explicitly exhibiting it.
It seems to be the same phenomenon in your case, where you have a computable function, but you don't know which algorithm computes it.
Addition. Let me try to address Thierry Zell's concern about the
third question. To my way of thinking, the phenomenon of
the question is an instance of the problem of uniformity
of algorithms, a pervasively considered issue in
computability theory.
To illustrate, consider the question of whether a given
program $p$ halts on input $0$ before another program $q$.
Let $f_p(q)=1$ if it does and otherwise $f_p(q)=0$. Every
such function $f_p$ is computable, for similar reasons to
my $\pi$ function $f$ above, since either $p$ doesn't halt
at all on input $0$, in which case $f_p$ is identically
$0$, or $p$ does halt in $N$ steps, in which case we need
only run $q$ for $N$ steps to see if it halts, and give our
output for $f_p(q)$ by that time. So each $f_p$ is a
computable function. But the joint function
$f(p,q)=f_p(q)$, a binary function, is not computable (if
it were, then we could solve the halting problem: to decide
if $p$ halts on input $0$, design a program $q$ that would
take one step extra after a halt, and ask if $p$ halts
before $q$).
In other words, the function $f(p,q)$ is computable for any
fixed $p$, but not uniformly in $p$. And such uniformity
issues are ubiquitous in computability theory.
In the example of the question, each class of graphs is
decidable, but not uniformly so, since by Tony's answer
there is no uniform algorithm, given a description of the
class, to find the collection of excluded minors. But for
any such fixed class, the membership question is decidable.
The issue of whether a given algorithm is uniform in a
given parameter is a very common concern in computability
theory, and occurs throughout the subject.
Best Answer
For a reference concerning this problem, see Cattell et al, "On computing graph minor obstruction sets", Theor. Comput. Sci. 2000.
I think the answer to your specific question is that it's recursively enumerable (one can test all graphs using P to see whether they belong to E) but not recursive (without further information there is no way of knowing that one has found all obstructions).
Here's a specific construction that shows this: given an instance h of the halting problem, construct an algorithm A that either recognizes all graphs if h is a non-halting instance, or that recognizes the graphs with no K_s minor if h halts in s steps. It's not hard to do this in such a way that A is always a polynomial time algorithm, but one can't tell whether E is empty or non-empty without solving the halting problem.