Are inclusions "canonical" injections?
The consensus seems to be that this is an answer to the question as stated (though I didn't originally realize that it was), so I'll go ahead and post it as one.
There are other ways to choose such a class of "good maps". For instance, you can transfer any class of good maps (such as the "standard" one consisting of inclusions) across an isomorphism of categories ${\rm Set} \cong {\rm Set}$ (though not necesarily across an equivalence of categories). To produce a nontrivial isomorphism ${\rm Set} \cong {\rm Set}$, choose two sets $X$ and $Y$ and an isomorphism $e:X≅Y$ that is not the identity. Define $F:Set→Set$ by $F(X)=Y$, $F(Y)=X$, and $F(A)=A$ for all other sets $A$, and define the action of $F$ on arrows by composing with the chosen isomorphism $e$ whenever needed. Then $F$ is an isomorphism ${\rm Set} \cong {\rm Set}$, under which a subset inclusion $i:X′↪X$ (for $X' \neq X,Y$) is sent to an injection $ei:X′→Y$ that is not (usually) an inclusion. More generally, you can construct an automorphism of ${\rm Set}$ that essentially arbitrarily permutes each bijection-equivalence-class of sets.
A way to make the question more interesting, as suggested by Peter Lumsdaine and Asaf Karagila, is to ask whether there is a class of good maps that is not related to the inclusions by such an automorphism of ${\rm Set}$, or similarly that is not equivalent as an M-category to the standard M-category of sets, functions, and inclusions. I don't know the answer to this version of the question.
The example Mike Shulman gives in the comments answers the question. Here is another example illustrating that more complicated things can happen as well: even identity maps need not be good morphisms by the OP's definition. (This cannot happen in an $M$-category or a system of inclusions, where all identity maps have to be inclusions.)
If $X=\operatorname{Hom}(\{\emptyset\},X')$ for some $X'$, let the good morphisms to $X$ be the inclusions. For all other $X$ let the good morphisms to $X$ be $e\circ i$, where $i:Z\to\operatorname{Hom}(\{\emptyset\},X)$ is an inclusion and $e$ is evaluation at $\emptyset$.
As raised in Asaf Karagila's comment, this does come from an endofunctor of $\operatorname{Set}$, namely $F(X)=X$ if $X=\operatorname{Hom}(\{\emptyset\},X')$ and $F(X)=\operatorname{Hom}(\{\emptyset\},X)$ otherwise (and do the obvious thing to morphisms). One could do something similar for any idempotent endofunctor.
If one is allowed to mess with the construction of products/limits then one can arrange for this to respect these operations. If one insists on constructing them in a (the?) standard way, then I'm not sure.
Edit: Here is a formulation of the "is it always the same M-category" question which allows for cases like this one and the one in user44191's answer. To a notion of good morphisms associate the following data. There is a functor $F:\operatorname{Set}\to\operatorname{Set}$ with $F\circ F=F$, and there is an isomorphism $f_X:F(X)\to X$ for each $X$, such that $f_{F(X)}=\operatorname{id}_{F(X)}$ and the elements of $P(X)$ are all morphisms $f_X \circ i $ for some injections $i: Z\to F(X)$.
To get hold of these, let $g: A\to X$ be the good morphism corresponding to $\operatorname{id}_X$, take $F(X)=A$ and $ f_X= g$. Properties 2, 1 and then 3 applied to $f_X\circ f_{F(X)}$ give $F\circ F=F$ and $f_{F(X)}=\operatorname{id}_{F(X)}$.
Is there always an automorphism of $\operatorname{Set}$ making the maps $i$ into inclusions (in the ordinary sense)?
For this post, I'm going to assume that the set-universe has a well-order on objects (including sets). This is true if your universe is the constructible universe (as such an order can be built inductively).
Then the class of all order-preserving injections from ordinals is a "good" class, which is not equivalent to the usual class of injections.
Property 1: for every injective map $f: Y \rightarrow X$, the image of $f$ is a subset of a well-ordered set ($X$; the well-ordering comes from the universal well-ordering), and so is well-ordered. It therefore has a unique bijection to some ordinal.
Property 2: The composition of an injection from an ordinal and any injection will be an injection from an ordinal.
Property 3: If $h: A \rightarrow C$ is an injection from an ordinal, then $f: A \rightarrow B$ must be an injection and start with an ordinal.
This is not equivalent to the original, if I've understood the proper idea of equivalence; very few sets act as domains for "good" maps for this class, while in the original, every set was the domain for a "good" map.
I think it may be better to look at classes of "good" maps instead as endofunctors $P: \text{Set} \rightarrow \text{Set}$. Given a class of "good" maps, define $i_X: P(X) \rightarrow X$ as the unique map and domain (under property 1) corresponding to $id_X: X \rightarrow X$. Then for any function $f:Y \rightarrow X$, define $P(f) = i_X \circ f \circ i_Y^{-1}$. This defines an endofunctor of $\text{Set}$.
I think this collapses the equivalent classes of "good" maps discussed in the commments, in that those two classes result in the same endofunctor, and that "goodness" may be possible to determine using the endofunctor itself.