What is the exact meaning of Lorentz invariance for a quantum scalar field? In axiomatic QFT, the defining property of a scalar field $\phi$ is that it does not change under a Lorentz transformation: that is, "If $U(\Lambda)$ is the unitary representation of a Lorentz boost $\Lambda$, we have $U(\Lambda)^\dagger \phi(x) U(\Lambda)=\phi(\Lambda^{-1} x)$." Call the quoted part Statement A.
In Lagrangian QFT, it is required that a scalar field must have a scalar Lagrangian, which in turn implies that "If $\phi(x)$ solves the equations of motion then so does the transformed field $\phi(\Lambda^{-1} x)$". Call the quoted part Statement B.
Does Statement A imply Statement B? I can't see how the transformation law leads to the requirement that the same equations of motion should hold in every inertial frame.
For reference, the full transformation law for the field, including a spacetime translation $a$, is given by $U(a,\Lambda)^\dagger \phi(x) U(a,\Lambda)=\phi(\Lambda^{-1}(x-a))$. I'd be grateful for an explanation in terms of this transformation law.
 A: Classical field theory
Let's start with classical field theory of a scalar field, $\phi(x^\mu)$, where $x^\mu$ are coordinates on spacetime. Then all dynamics are derivable from the action via the Euler-Lagrange equations. The action typically takes the form
\begin{equation}
S = \int {\rm d}^4 x \mathcal{L}(\phi,\partial\phi)
\end{equation}
Under a Lorentz transformation $\Lambda$, $\phi(x)\rightarrow\phi(\Lambda^{-1}x)$. If the action remains invariant when we perform this transformation, then the equations of motion will be invariant under this transformation. As a result, if $\phi(x)$ obeys the equations of motion, then so will $\phi(\Lambda^{-1}x)$.
Crucially, we can apply a Lorentz transformation to $\phi$, whether or not the action is invariant under a Lorentz transformation. The notion of invariance, or symmetry, has two parts: (1) we apply the transformation, and (2) the action does not change when we perform the transformation.
An example of an action which will be invariant under Lorentz transformations is
\begin{equation}
S = \int {\rm d^4 x} \left(-\frac{1}{2} \eta^{\mu\nu}\partial_\mu \phi \partial_\nu \phi - V(\phi)\right)
\end{equation}
where I've chosen the metric signature $\{-1,+1,+1,+1\}$. Meanwhile, an action which is not Lorentz invariant is
\begin{equation}
S = \int {\rm d^4 x} \left(-\frac{1}{2} \eta^{\mu\nu}\partial_\mu \phi \partial_\nu \phi + V(x)^\mu \partial_\mu \phi)\right)
\end{equation}
where $V(x)^\mu$ is some 4-vector that depends on space. For example, we could take $V(x)^\mu=x^\mu$. To check the action is not Lorentz invarant, we can pefrom the transformation $\phi(x)\rightarrow\phi(\Lambda^{-1}x)$ and simply check that the action changes. This shows that the transformation law, by itself, is not enough to guarantee the theory is invarant.
Quantum field theory (canonical formulation)
The story is similar, but richer, at the quantum level. If we take a path integral approach, then the above story is (almost) sufficient quantum mechanically (with the subtlety being that we have to check if the path integral measure is also invariant, after renormalization).
In the canonical formalism, $\phi(x)$ is an operator valued distribution on spacetime. The equation $U(\Lambda)^\dagger \phi(x) U(\lambda) = \phi(\Lambda^{-1}x)$ explains how to represent Lorentz transformations of the field on Hilbert space.
In "vanilla" non-relativisitc quantum mechanics, to check if a given transformation $T$ is a symmetry, we represent the transformation $T$ via unitary operators $U(T)$ and see if the Hamiltonian is invariant, $U^\dagger(S)HU(S)=H$.
In our relativistic case, we have to be more careful, since boosts will change the Hamilonian. What we really need is that the Lorentz transformations act on the energy-momentum 4-vector as
\begin{equation}
U^\dagger(\Lambda) P^\mu U(\Lambda) = \Lambda^\mu_{\ \ \nu}P^\nu
\end{equation}
where $P^\mu=\{H,P^i\}$, and $H$ is the Hamiltonian and $P^i$ are the spatial components of the momentum. We also require that the angular momentum tensor $M^{\mu\nu}$ transforms as a tensor
\begin{equation}
U^\dagger(\Lambda) M^{\mu\nu} U(\Lambda) = \Lambda^\mu_{\ \ \alpha} \Lambda^\nu_{\ \ \beta} M^{\alpha\beta}
\end{equation}
The quantities $P^\mu$ and $M^{\mu\nu}$ can be derived by using Noether's theorem, applied the action, for spacetime transformations and Lorentz transformations, respectively. Explicit forms for the scalar field can be found, for example, in Chapter 1 of David Tong's QFT lecture notes, http://www.damtp.cam.ac.uk/user/tong/qft.html.
In practice, one often works with an infintesimal version of these laws. For a translation, we write $U(\Lambda)=1+i a^\mu P_\mu$, where $a^\mu$ is the 4 vector defining the translation, and $P_\mu$ is the "generator" of the transformation, which we identify with the Hamiltonian and momentum as above. For a Lorentz transformation (boosts and rotations), $U(\Lambda)=1+i \omega^{\mu\nu} M_{\mu\nu}$, where $\omega^{\mu\nu}=-\omega^{\nu\mu}$ are the parameters of the Lorentz transformation, and $M_{\mu\nu}$ are the generators. Then the group transformation laws above, imply the following commutation rules for the generators
\begin{eqnarray}
[P_\mu,P_\nu]&=&0 \\
[M_{\mu\nu},P_\rho]&=& -i \left( \eta_{\mu\rho} P_\nu - \eta_{\nu \rho} P_\mu\right)\\
[M_{\mu\nu},M_{\rho\sigma}]&=&-i\left(\eta_{\mu \rho}M_{\nu\sigma} - \eta_{\mu \sigma}M_{\nu\rho}-\eta_{\nu\rho}M_{\mu\sigma}+\eta_{\nu\sigma}M_{\mu\rho}\right) \\
\end{eqnarray}
So, the steps for checking whether a given field theory has a Lorentz symmetry in canonical quantization are:

*

*Write down the action

*Pass to the Hamiltonian framework by using a Legendre
transformation

*In the process of passing to the Hamiltonian framework, derive
commutation relations for the fields

*Construct the 4-momentum and angular momentum operators

*Check that the operators from step 4 obey the Poincair{'e}
algebra (set of commutation relations above).

A few additional notes:

*

*Often, if the action is Lorentz invariant, then the quantum theory will be as well.


*Sometimes this logic is inverted and we start from the commutation relations and construct a theory which obeys the necessary relationships.


*In classical field theory, we can formulate the question of symmetry in a very similar way using Poisson brackets.


*The approach in this answer was to find a generalization of the idea from non-relativistic quantum mechanics that "a symmetry leaves the Hamiltonian invariant."  However, we can take a more abstract point of view as well, in that we want to have a unitary representation of the Poincair{'e} group on Hilbert space. The Noether charges $P_\mu$ and $M_{\mu\nu}$ provide the generators for this representation.
To come back to your question: the Lorentz transformation law, by itself, does not imply the theory has Lorentz symmetry. However, it is an important ingredient in checking if a given theory is Lorentz invariant. And, the fact that we were able to construct a unitary representation of the Poincair{'e} group on Hilbert space, is equivalent to having the theory being symmetric.
A: 
What is the exact meaning of Lorentz invariance for a quantum scalar field?

The question proposes this definition (statement B):

If $\phi(x)$ solves the equations of motion then so does the transformed field $\phi(\Lambda^{-1}x)$.

That definition can either work or fail, depending on how we write the equation of motion! I'll show two different ways of writing the equation of motion: one where statement B works, and one where it fails. Then I'll recommend a different definition of symmetry that works fine no matter how we write the equation of motion.
 Two ways of writing the equation of motion 
Suppose that we have a unitary representation of the Poincaré group whose members $U(a,\Lambda)$ satisfy
$$
 U^{-1}(a,\Lambda)\phi(x)U(a,\Lambda)=\phi(\Lambda^{-1}(x-a)).
\tag{A}
$$
This is statement A extended to the whole Poincaré group. This implies the equation of motion, because it dictates the time-dependence of $\phi(x)$. Setting $\Lambda=1$ gives
$$
 U^{-1}(a,1)\phi(x)U(a,1)=\phi(x-a),
\tag{1}
$$
which is one way of writing the equation of motion. The case $a=(t,0,0,0)$ describes the time-dependence of $\phi(x)$. By taking derivatives of (1) with respect to the components of $a$ and then setting $a=0$, we can derive
$$
\newcommand{\pl}{\partial}
 \pl^2\phi(x)+\big[P^\mu,[P_\mu,\phi(x)]\big] = 0
\tag{2}
$$
where $P_\mu$ are the generators of $U(a,1)$. This is another way of writing the equation of motion.
Interestingly, statement B works when applied to (2), but it fails when applied to (1). The proofs are shown below.
 Statement B works when applied to (2) 
Define a transformed field
$$
 \tilde\phi(x)
 \equiv 
 \phi(\Lambda^{-1}x).
\tag{3}
$$
Using the abbreviation $x' \equiv \Lambda^{-1}x$, the proof is easy:
\begin{align}
 \pl^2\tilde\phi(x)+\big[P^\mu,[P_\mu,\tilde\phi(x)]\big]
 &=
 \pl^2\phi(x')+\big[P^\mu,[P_\mu,\phi(x')]\big]
 \\
 &=
 (\pl^2\phi)(x')+\big[P^\mu,[P_\mu,\phi(x')]\big].
\tag{4}
\end{align}
On the last line, $\pl^2\phi$ denotes the derivatives with respect to the argument of $\phi$, whatever that argument is. The last line follows from the first line because the differential operator $\partial^2$ is Lorentz-invariant. Requiring that this quantity be zero for all spacetime points $x'$ is the same as requiring equation (2) for all spacetime points $x$, so statement B works in this case.
 Statement B fails when applied to (1) 
Define $\tilde\phi(x)$ as before and consider the identities
\begin{align}
 U^{-1}(a,1)\tilde\phi(x)U(a,1)
 &=
 U^{-1}(a,1)\phi(\Lambda^{-1}x)U(a,1)
\\
 &=\phi(\Lambda^{-1}x-a)
\tag{5}
\end{align}
and
\begin{align}
 \tilde\phi(x-a)
 &=\phi(\Lambda^{-1}(x-a))
\\
 &=\phi(\Lambda^{-1}x-\Lambda^{-1}a).
\tag{6}
\end{align}
Typically,
$$
 \Lambda^{-1}a\neq a,
\tag{7}
$$
so (5) and (6) are typically not equal to each other. In other words, the transformed field $\tilde\phi(x)$ does not satisfy the equation of motion (1).
Altogether, this shows that statement B depends on how the equation of motion is written. Andrew's good answer said the same thing in a different way: in general, for statement B to work as desired, we would also need to transform the generators of $U(a,1)$.
 Recommended definition 
Let $A(R)$ be the set of observables localized in a spacetime region $R$. In other words, $A(R)$ is the set of observables that can be constructed algebraically from $\phi(x)$ with $x\in R$. If a unitary transformation $U$ satisfies $U^{-1}A(R)U=A(R')$ for some spacetime isometry $x\to x'$ that is the same for all regions $R$, then we call it a symmetry. If $x'=x$, then we call it an internal symmetry.$^\dagger$
$^\dagger$  The spacetime regions $R$ can be arbitrarily small. The definition refers to regions, instead of points, to avoid the technical problems that occur when trying to define operators localized at individual points. 
With this recommended definition of symmetry, statement A in the question implies that the theory is Lorentz symmetric. This is clear by inspection.
Notice that this recommended definition describes the effect of $U$ for all times, so it implicitly says that $U$ must be consistent with the theory's dynamics. It's consistent with the intent of statement B, but it doesn't depend on how we write the equation of motion.
