The writing is certainly less clear than it ought to be.
Your instincts are right; they just didn't explain it well.
In particular, they're not saying that the term in brackets
only vanishes for symmetry transformations. It always
vanishes for a field obeying the Euler-Lagrange equations.
What they're really saying is:
- A continuous transformation can be written as equation (2.9)
- The equation of motion will be invariant under that continuous transformation if the Lagrangian changes as equation (2.10)
- Let's see how the Lagrangian actually changes under the transformation of (2.9)
- It changes as a term that looks like $\alpha \partial_\mu \mathcal{J}^\mu$ plus a term that goes to zero because of the Euler-Lagrange equations for the original, un-transformed system
- The actual change we get is the change we demanded for the Lagrangian to be invariant up to a 4-divergence
- Therefore, the continuous transformation is a symmetry transformation
The text immediately following what the OP included says
The second term [in equation (2.11)] vanishes by the
Euler-Lagrange equation. We set the remaining term equal to
$\alpha \partial_\mu \mathcal{J}^\mu$ and find
\begin{equation}
\partial_\mu j^\mu(x) = 0\,
\qquad \mathrm{for} \qquad
j^\mu(x) = \frac{\partial \mathcal{L}} {\partial
(\partial_\mu \phi)} \Delta \phi - \mathcal{J}^\mu.
\end{equation}
This really makes it look like they're saying that $j^\mu$ is identically zero; in fact, I would argue that that's the most reasonable
reading of the text. What they should have said is that
we get the generic result
\begin{equation}
\Delta \mathcal{L} = \partial_\mu \left( \frac{\partial
\mathcal{L}} {\partial (\partial_\mu \phi)} \Delta \phi
\right) \tag{A}
\end{equation}
whenever $\phi$ satisfies the Euler-Lagrange equations, but
to call $\Delta \phi$ a symmetry transformation we require
the existence of some other field $\mathcal{J}^\mu$ such
that
\begin{equation}
\Delta \mathcal{L} = \partial_\mu \mathcal{J}^\mu \tag{B}
\end{equation}
regardless of whether or not the Euler-Lagrange equations
are satisfied. Again, (A) is generically true for any
differentiable Lagrangian and field satisfying the E-L
equations. On the other hand, you have to use the
particular form of $\mathcal{L}$ and the particular form of
$\Delta \phi$ to decide if (B) is true in any particular
case. (Also note that $\mathcal{J}^\mu = 0$ is perfectly
acceptable.)
It probably would have been clearer if P&S had begun by
moving the explanation of what a symmetry transformation is
[the four sentences up to equation (2.10)] to a previous
paragraph and expanding on it. Then they could introduce a
continuous transformation and show how the Lagrangian
changes. But why bother editing a piece of text for
clarity, when it's only explaining something as frivolous
and useless as Noether's Theorem? ;)
These issues are made much clearer in @joshphysics's version of the
proof — in particular with his introduction of separate
$\mathcal{J}^\mu$ and $\mathcal{K}^\mu$ vector fields. They
look like they perform the same function, and yet they're
different, as he explains in the "Important Notes"
section
at the bottom of his answer, and in some of the
comments.
I also highly recommend the
excellent "Quantum Field Theory for the Gifted Amateur", which does
a typically superb job on Noether's Theorem, and is really what I think should be
the standard text. The title is deceptive because it's
aimed at professional physicists, just not current
professional quantum-field theorists. I encourage you to at
least read the other
answer, or
even take up "Gifted Amateur" if you really want to
understand QFT.
