Counting massive degrees of freedom after gauge fixing

What you can remove with a field redefinition (of the form of a $U(1)$ gauge transformation) is the phase of $\phi$. But the total number of degrees of freedom (dof's) is not going to change since the resulting lagrangian is not longer gauge invariant. The new counting is $3+1$ where the $3$ dof's come from $A_\mu$ (with a covariant constraint, see below, so that really $3=4-1$), and the 1 dof comes instead from the scalar radial mode of $\phi$.

More explicitly, write $$\phi(x)=e^{ie\pi(x)}\frac{h(x)}{\sqrt{2}}$$ (with both $\pi$ and $h(x)$ real scalar fields) and the covariant derivative $D_\mu \phi=(\partial_\mu-i e A_\mu)\phi$ becomes $$ D_\mu \phi=e^{ie\pi}\left[ie(\partial_\mu\pi-A_\mu)+\frac{1}{\sqrt{2}}\partial_\mu h\right] $$ so that the lagrangian reads $$ \mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{e^2}{2}h^2(\partial_\mu\pi-A_\mu)^2+\frac{1}{2}(\partial_\mu h)^2-\frac{m^2}{2}h^2\,. $$ Now, defining the new variable $A_\mu^\prime=A_\mu-\partial_\mu\pi$ (which is the gauge invariant combination), the lagrangian becomes $$ \mathcal{L}=-\frac{1}{4}F^\prime_{\mu\nu}F^{\prime \mu\nu}+\frac{e^2}{2}h^2 A_\mu^{\prime 2}+\frac{1}{2}(\partial_\mu h)^2-\frac{m^2}{2}h^2 $$ in full analogy to the abelian Higgs mechanism except that $h$ has vanishing vacuum expectation value. Now, this last lagrangian depends on the fields $A_\mu^\prime$ and $h$: how many dof's are there? Well, the $h(x)$ certainly counts 1. The $A_\mu^\prime$ on the other hand counts $3$ and neither $4$ (even though $\mu=0,1,2,3$) nor $2$. Indeed, from the equations of motion $\partial_\mu F^{^\prime\mu\nu}=-e^2h^2 A^{\prime \nu}$ we see the covariant constrain $$ \partial_\mu (A^{\prime\mu} h^2)=0 $$ which sends us from $4$ to $3$. But notice that there is no gauge invariance for $A_\mu^\prime$ in its lagrangian above (the $e^2 h^2 A_\mu^{\prime 2}$-term breaks it) and no longitudinal mode can be therefore removed in that way: $A_\mu^\prime$ is a different physical configuration than say $A^\prime_\mu-\partial_\mu \Omega$ (in particular one solves the equation of motion, the other doesn't). So the counting ends here and it matches that for massive spin-1 plus a real scalar, that is $3+1$ as expected. Notice however, that the spin-1 particle is not massive and all this gymnastic is just artificial as you wanted to move around one scalar dof $\pi$ inside $A_\mu$.

Before we can compare with the second Lagrangian (2), the first Lagrangian (1) should include a gauge-fixing term ${\cal L}_{\rm gf}$, e.g. ${\cal L}_{\rm gf} = \lambda~ {\rm Im}(\phi),$ where $\lambda$ is a Lagrange multiplier. After integrating out $\lambda$ and ${\rm Im}(\phi)$ the Lagrangian (1) becomes the Lagrangian (2).

Why we need to consider gauge-fixed (rather than un-gauge-fixed) Lagrangians is e.g. discussed in my Phys.SE answer here.

For both Lagrangians, the scalar field effectively induces a mass term for the $A_{\mu}$-field.

$\downarrow$ Table 1: Real DOF of OP's Lagrangians.

$$\begin{array}{ccc} \text{Lagrangian}& \text{Off-shell DOF}^1 & \text{On-shell DOF}^2 \cr (1)& 2+4-1=5 &2+3-1=4 \cr (2)& 1+4-0=5 &1+3-0=4 \end{array}$$

$^1$ Off-shell DOF = # (components)- # (gauge transformations).

$^2$ On-shell DOF = # (helicity states)= (Classical DOF)/2, where Classical DOF = #(initial conditions).