You have to be careful about what you mean by $\text{PA}^-$. The convention I expect is that $\text{PA}^-$ is the finite set of axioms for a discretely ordered semiring - which includes the axiom that multiplication is commutative. This set of axioms is typical when we want to look at systems of arithmetic with limited induction. For example, it is the set of axioms in Kaye's book The Structure of Models of Peano Arithmetic.
The original five axioms proposed by Peano, which only mention the successor operation, but not addition, multiplication, or order, are not sufficient for first-order Peano Arithmetic, because they are not sufficient to define addition or multiplication in first-order semantics.
On the other hand, some authors use an abbreviated set of axioms for Peano arithmetic. For example, Mendelson's logic text uses a smaller set of axioms that do not mention the order relation. These work because he only considers them in conjunction with the induction scheme. It seems to me that $M_2(\mathbb{N})$ is a model of this smaller set of axioms, but I would not call them $\text{PA}^-$.
Even when we look at the full set of axioms for $\text{PA}^-$, we can give an example of a model that does not satisfy induction. Tennenbaum's theorem shows that no nonstandard model of Peano arithmetic is computable. But the set of polynomials over $\mathbb{Z}$ with positive leading coefficient are a computable nonstandard model of $\text{PA}^-$, so induction must fail in that model, although I am not sure which instance of the schema fails. I believe the proof only relies on a finite number of instances of induction, so that would give a short list of candidates.
1) Define $n+0=n$. Assume $n+m$ has been defined. Then define $n+\sigma(m)=\sigma(n+m)$. This defines addition on all ordered pairs of naturals. Suppose there was some alternate addition $+'$ satisfying the above properties. We notice that for pairs $(n, 0)$, $n+0=n+'0$. Assume that for each $n$, we have shown that $n+k=n+'k$ for all $0 \leq k \leq m$. Then $n+\sigma(k)=\sigma(n+k)=\sigma(n+'k)=n+'k$ and so we see that $+, +'$ are the same for all pairs. This gives uniqueness. You will want to prove distributivity of addition for later parts. Try to get this via induction.
2) Pick any $n \in \mathbb{N}$. We proceed by induction on $m$. If $m=0$, then $p=m$. If there exists $p$ so that $n+p=m$, then we observe that $n+ \sigma(p)=\sigma(n+p)=\sigma(m)$. If there exists $p$ so that $n=p+m$, then if $p=0$, we have $n+1=\sigma(m)$. If $p \neq 0$, then the number we need is the natural whose successor is $p$. Try proving that if $p \neq 0$, it has a predecessor. (Hint: induction!)
3) This relation is clearly reflexive. For anti-symmetry, if there exists $p$ so that $a+p=b$ and $p'$ so that $b+p'=a$, then $b+p+p'=b$. You should be able to show this implies $p+p'=0$. If $p \neq 0$ or $p' \neq 0$, this should give you a predecessor of $0$, which is a contradiction and shows that $p=p'=0$ and so $a=a+p=b$. Transitivity is easy, and totality is part 2).
4) This is not hard. Try to work it out for yourself.
5) Your intuition is correct. You can easily show that $\sigma (n) \geq n$ but $n \neq \sigma(n)$.
6) If this were not the case, you would have an infinite descending chain in $\mathbb{N}$. To get a contradiction, you need to show that for each $m \in \mathbb{N}$, there are only finitely many naturals less than $m$. You can do this by induction.
7) You are correct.
8) This is somewhat similar to addition. Define inductively and use your definition to prove what you need.
9) You can do this by induction.
Best Answer
This is how I would go about it, in three steps.
Prove $0=m\cdot 0=0\cdot m$ for all $m\in\omega$. As you said, you can easily show this.
Prove $m'n=mn+n$ for all $m,n\in\omega$. We can do this by induction. Let $$ K=\{n\in\omega\ |\ m'n=mn+n\} $$ By definition, $m'\cdot 0=0$, and $m\cdot 0+0=0+0=0$, so $0\in K$. Suppose $n\in K$. Then $$ m'n'=m'+m'n=m'+mn+n=m+mn+n'=mn'+n' $$ where I have used the second facts you listed for addition and multiplication, and I assume you know $a'+b=(a+b)'=a+b'$, which is usually used in proving the commutativity of addition. So $n'\in K$.
We can now prove $mn=nm$ for all $m,n\in\omega$. Let $S=\{m\in\omega\ |\forall_{n\in\omega}\ mn=nm\}$. By Step 1, $0\in S$. Let $m\in S$. Then $$ m'n=mn+n=nm+n=n+nm=nm' $$ so $m'\in S$, so $S$ is inductive, and $S=\omega$.