To prove this, use naturality of the Leray-Serre spectral sequence: the $G$-equivariant map $\ast \to M$ induces a map of spectral sequences $$ H^p(G;H^q(M;\Bbb Z)) \to H^p(G;H^q(\ast;\Bbb Z)). $$ This map is compatible with the differentials, the map is an isomorphism when $p=0$, and the target is zero when $p > 0$. If an element $y$ on the 0-line were in the image of a differential, $y = d_r(x)$, then applying this map of spectral sequences we'd find $y = d_r(0)$ and hence $y=0$.

To prove this, use naturality of the Leray-Serre spectral sequence: the fixed point has a $G$-equivariant map $\ast \to M$ that induces a map of spectral sequences $$ H^p(G;H^q(M;\Bbb Z)) \to H^p(G;H^q(\ast;\Bbb Z)). $$ (This converges to the map $H^*_G(M) \to H^*_G(\ast)$.)

This map is compatible with the differentials, the map is an isomorphism when $p=0$, and the target is zero when $p > 0$. If an element $y$ on the 0-line were in the image of a differential, $y = d_r(x)$, then applying this map of spectral sequences we'd find $y = d_r(0)$ and hence $y=0$.

This is false; the differentials do not necessarily have image zero.

The problem comes earlier when you want to define the "reduced" groups: the map $H^p_G(M;\Bbb Z) \to H^p_G(\ast;\Bbb Z)$ is not necessarily surjective.

A standard example would be the case where $G = C_2$ is the cyclic group of order two and $M = S^n$ is the n-sphere with the antipodal action of $G$. In that case, $EG \times_\phi M$ is homotopy equivalent to the real projective space $\Bbb{RP}^n$, and in particular its integer cohomology vanishes in degrees greater than $n$. However, $$E_2^{p,0} = H^p_{C_2}(\ast;\Bbb Z) \cong H^p(BC_2;\Bbb Z)$$ is nontrivial for all even values of $p$.

To prove this, use naturality of the Leray-Serre spectral sequence: the $G$-equivariant map $\ast \to M$ induces a map of spectral sequences $$ H^p(G;H^q(M;\Bbb Z)) \to H^p(G;H^q(\ast;\Bbb Z)). $$ This map is compatible with the differentials, the map is an isomorphism when $p=0$, and the target is zero when $p > 0$. If an element $y$ on the 0-line were in the image of a differential, $y = d_r(x)$, then applying this map of spectral sequences we'd find $y = d_r(0)$ and hence $y=0$.