I cannot see how follows the equality $\cal M^{\bot}=M^*$-Inj below: (both inclusions are wanted)
where
$X \in {\cal H}^\bot \iff \forall f\in{\cal H}: \mathrm{Hom}(f,X)$ bijective
and
$X \in {\cal H}$-Inj $\iff \forall f\in{\cal H}: \mathrm{Hom}(f,X)$ surjective
It is actually quite straightforward:
($\subseteq$) Let $X\in {\cal M}^\bot$. Because ${\cal M}^\bot \subseteq {\cal M}$-Inj, we only need to check that $X$ is injective w.r.t. all the $m^*$. So let $f\colon A^* \to X$.
(1) the equation $f\circ p_1\circ m = f\circ p_2\circ m$ forces $f\circ p_1 = f\circ p_2$ because both maps extend $f\circ p_1\circ m$ along $m$.
(2) Let $h=f\circ p_1=f\circ p_2$. For $i=1,2$ the equations $h \circ m^* \circ p_i = h = f \circ p_i$ force $h \circ m^* = f$ because of the universal property of the pushout.
Therefore $h$ indeed extends $f$ along $m^*$.
($\supseteq$) Let $X\in {\cal M}^*$-Inj. Given $f\colon A \to X$, we already know that $f$ can be extended along $m$ because $X\in {\cal M}$-Inj. So we only need to check that such an extension is unique.
Suppose $h_1,h_2\colon A\to X$ are two maps with $q_1\circ m = f = q_2\circ m$. Then the pushout gives a unique map $g\colon A^*\to X$ with $g\circ p_i = q_i$ ($i=1,2$). Let $h\colon A'\to X$ be an extension of $g$ along $m^*$, i.e. $h\circ m^* = g$. Then
$$q_1 = g \circ p_1 = h\circ m^* \circ p_1 = h = h\circ m^* \circ p_2 = g \circ p_2 = q_2$$