Your proof of normality, definition of $\phi$ and proof that $\phi$ is well-defined are spot on (though I'd personally write something like "let $H = \ker(\varphi)$" at the start to avoid writing $\ker(\varphi)$ everywhere, as it makes things look slightly cleaner). Also, your proof that the map is a homomorphism is correct.
As for your starred line, as you're showing injectivity, I'd suggest replacing it with a simpler argument. In particular, as you have a homomorphism, to show injectivity it suffices to show that the kernel is trivial. But if $gH \in \ker(\phi)$, then by definition $\varphi(g) = 1$, so $g \in H$ and $gH = H$, which is the identity element in the quotient group.
An alternative proof - which implicitly proves the claim about trivial kernel that I mentioned above - is to say that if $\phi(gH) = \phi(hH)$, then $\varphi(g) = \varphi(h)$, so that $\varphi(gh^{-1}) = 1$ and $gh^{-1} \in H$. Thus $gH = hH$, as required.
I don't entirely follow your argument at $(*)$; what is the $\pi$ that crops up in the penultimate expression? Also, do you mean to write $\phi^{-1}\varphi[h\ker(\varphi)]$ in the second expression? My guess is that $\pi$ is the projection map from $G$ to $G/\ker(\varphi)$, and that in the end you're slightly assuming what you're trying to prove.
As an example, of a case where this might not prove injectivity, let's suppose that $J$ and $H$ are normal subgroups of $G$ with $J < H$ (strict containment). Then if $H$ is the kernel of some homomorphism $\varphi$ out of $G$, then we consider the map
$$\phi: G/J \longrightarrow \varphi(G)$$
given by
$$\phi(gJ) = \varphi(g).$$
In particular, this isn't an injective map. Then following your line $(*)$, we say that
$$\phi^{-1}(hJ) = \phi^{-1}\varphi[hJ] = \phi^{-1}\varphi(\pi^{-1}\{hJ\}),$$
where $\pi$ is the projection $G \rightarrow G/J$. Does your argument then prove injectivity in this case? As if it does, then there's a mistake somewhere!
(Alternatively, you might take $\pi$ to be the projection map $G \rightarrow G/\ker(\varphi) = G/H$. But in this case you can't work with it unless you know exactly what $\ker(\varphi)$ is, in which case you already know if it's injective or not!)
Proposition $2$ is certainly true as soon as the operation on $G/H$ is well-defined, i.e. as soon as it doesn't depend on the choice of representatives $a$ and $b$ for the cosets $aH$ and $bH$. This is the case if $H$ is a "normal submonoid" as in your Definition 2, and can be shown more or less like in groups.
But it is not true that $aN=N$ is equivalent to $a\in N$ if $N$ is a submonoid: for example if you take the monoid of natural numbers with operation given by multiplication, then $N=\{0,1\}$ is a submonoid, and $0\in N$; but $0N=\{0\}\neq N$.
As I mentionned in a comment, the true problem is that the kernel of a homomorphism can be trivial without the homomorphism being injective. For example consider the homomorphism
$$\varphi : (\Bbb N,\cdot ,1)\to (\Bbb N,\cdot ,1): x\mapsto 0 \text{ iff } x\neq 1;$$only $1$ is sent to $1$, so the kernel is trivial; but it is not injective, since every natural other than $1$ is mapped to $0$. Thus the quotient monoid (if it is well-defined) should be isomorphic to $\Bbb N$, while the image of $\varphi$ is the submonoid $N$ defined above. The problem with monoids is that unlike in groups, you can't take inverses, so the kernel doesn't give you enough information.
There is however a form of isomorphism theorem that holds; instead of quotienting by a subobject, you need to quotient by the equivalence relation defined by
$$x\sim y\Leftrightarrow \varphi(x)=\varphi(y).$$
Then you can define an operation on the equivalence classes by putting $\overline{x}\cdot \overline{y}=\overline{x\cdot y}$, and show that $\varphi$ factorise through the quotient; and this factorisation will be injective because of the choice of the relation.
Best Answer
Checking that $\varphi$ is well-defined is indeed a crucial step of the proof (if you have seen a proof that omits this step, that is a serious gap!). So, suppose $g_1K=g_2K$, so $g_1^{-1}g_2\in K$. Since $K=\ker(f)$, this means $f(g_1^{-1}g_2)=1$. Since $f$ is a homomorphism, this means $f(g_1)^{-1}f(g_2)=1$ so $f(g_1)=f(g_2)$, and thus $\varphi$ gives the same value whether you use $g_1$ or $g_1$.