Let T1 and T2 be two Turing machines which enumerate recursively
enumerable sets S1 and S2 of integers which are disjoint but not
recursively separable (that is, there is no recursive set which
contains one and is disjoint from the other).  We also need the fact
that Peano arithmetic proves that S1 and S2 are disjoint (or, more
accurately, that there is no integer x which is enumerated by T1 and
by T2): this is harmless.  (The existence of T1 and T2 is proved,
e.g., in Roger's *Theory of Recursive Functions and Effective
Computability*, circa §7.7.)

Now given an integer x the statement "T2 never generates x" (that is,
"x does not belong to S2") is Pi_1.  Consider the set C of x such that
this is a theorem of Peano: if the set of Pi_1 theorems of Peano
arithmetic were recursive, then C would be recursive.  Certainly, C is
disjoint from S2 because Peano is sound.  But conversely, if x belongs
to S1 then that fact is a theorem in Peano (because it is a Delta_0
statement), so it is also a theorem in Peano that T2 never generates x
(because by assumption Peano proves disjointness of S1 and S2), i.e.,
x belongs to C.  So we have a recursive set C which contains S1 and is
disjoint from S2, a contradiction.

In fact the reasoning above shows that the set of Pi_1 theorems of
Peano and the set of Pi_1 antitheorems (=negation is a theorem) are
recursively inseparable.  So it is not possible that for all Pi_1
statements P either Peano+Consis(Peano) proves Consis(Peano+P) or
Peano+Consis(Peano) proves Consis(Peano + not-P): for if this were the
case we would have a recursive separation of the set of Pi_1 theorems
into certain consistent statements (including all Pi_1 theorems) and
certain unprovable statements (including all Pi_1 antitheorems).  In
particular, there is a Pi_1 statement P such that Peano+Consis(Peano)
does not prove either Consis(Peano+P) or Consis(Peano + not-P); and of
course, P must be true, because otherwise not-P would be a theorem (so
Peano+Consis(Peano) would prove Consis(Peano + not-P)), but it must
also be unprovable in Peano (otherwise Peano+Consis(Peano) would prove
Consis(Peano+P)).  Ergo, I have shown that there is a true Pi_1
statement P which is not provable in Peano (nor refutable) but such
that Peano+Consis(Peano) cannot show that P is unprovable nor can it
show that it is consistent.

And I'm sure P can be made explicit (using effective inseparability).
Also, I'm sure we can go much further (for all integers k, find a true
Pi_1 statement P such that any addition of up to k levels of "is
consistent" or "is unprovable" is true: P is true, P is consistent, P
is unprovable, "P is consistent" is consistent, "P is consistent" is
unprovable, "P is unprovable" is consistent, "P is unprovable" is
unprovable, etc., where each level refers to a formal system which
adds consistency of the previous one to its axioms).

I hope I didn't make any major mistakes here.

I'd still like to find a textbook reference to such questions, because
I'm sure there are more interesting things to be said than my modest
knowledge will let me discover.