This is exercise 1.1(1) of Serre's Arbres, amalgames, SL$_2$. I explain what I mean by the title of the question in the last paragraph.
Let $f_1:A\to G_1$ and $f_2:A\to G_2$ be two group homomorphisms (these are the "system morphisms" I'm talking about in the title of the question), and let $G$ be the amalgamated free product $G_1 *_A G_2$, which is defined as the pushout of the diagram $$G_1 \xleftarrow{f_1} A \xrightarrow{f_2} G_2.$$
Recursively define $A^1=G_1^1=G_2^1={1}$; $A^n$ the normal subgroup in $A$ generated by $f_1^{-1}(G_1^{n-1})$ and $f_2^{-1}(G_2^{n-1})$; and $G_i^n$ the normal subgroup in $G_i$ generated by $f_i(A^n)$ (for $i=1,2$). It's quite easy to show that $A^n\subseteq A^{n+1}$ and that $G_i^{n}\subseteq G_i^{n+1}$ and therefore that $A^\infty$ and $G_i^\infty$, defined as the union of the obvious subgroups, are normal subgroups of $A$ and $G_i$ respectively.
Let $\pi_i:G_i\to G_i/G_i^\infty$ be the canonical projections. Notice that $f_i^{-1}(G_i^\infty)=A^\infty$ which is therefore equal to the kernel of $\pi_i\circ f_i$. Hence this last morphism descends to an injection $f_i:A/A^\infty\to G_i/G_i^\infty$ (I don't bother changing its name).
Here's where I'm stuck. I need to show that $G$ is isomorphic to $G'=(G_1/G_1^\infty)*_{A/A^\infty}(G_2/G_2^\infty)$, and preferably without using any structure theorem (i.e. I think it should be possible to do it using the universal property of the pushout). Is there a quick way to prove this? Obviously by the aforementioned universal property there is a unique group homomorphism $\varphi:G\to G'$ such that $\varphi$ factorizes the cone made with the projections, etc, through the pushout diagram. But I'm not succeeding in showing that it's a bijection.
Note that were this proven, it would mean that in the algebraic study of amalgamated free products, we could always suppose that $f_1$ and $f_2$ are injective without any loss of generality. I hope this explains my rather cryptic question title.
I'm not sure if this is what you're looking for, but you can prove by induction that the canonical homomorphisms $\psi_i\colon G_i\to G$ have $G_i^n\leq \ker(\psi_i)$ for each $n$, and hence $G_i^\infty\leq \ker(\psi_i)$. This gives homomorphisms $G_i/G_i^\infty\to G$, and combining these with the universal property gives a homomorphism $G'\to G$ which is the inverse of the homomorphism $\varphi$ described in your post.