How to Apply Elementary Axioms from Kleene Star to an Inequality
I assume that in your definition $a \leq b$ iff $a + b = b$.

First, note that if $a \leq b$ and $b \leq a$, then $a = a + b = b + a = b$.
Therefore, in order to show that $a = b$, it is sufficient to show that $a \leq b$ and $b \leq a$.

Now, you want to show that $(a + b)^\ast = (a + ab + b)^\ast$. As explained in the previous paragraph, it is sufficient to show that inequalities with $\leq$ in both directions hold ("[...] for them to be equal the sets should be contained in each other").



The inequality $(a + b)^\ast \leq (a + ab + b)^\ast$ is almost trivial:

$$(a + b) + ab = a + ab + b $$

therefore $(a + b) \leq (a + ab + b)$, and by monotonicity (your ER-3), it thus holds 

$$(a + b)^\ast \leq (a + ab + b)^\ast.$$



The other direction is a bit trickier.

By monotonicity (ER-2), (def-$\leq$), (Ax-1):

$$a(a + b)^\ast \leq (a + b)(a + b)^\ast \leq 1 + (a + b)(a + b)^\ast \leq (a+b)^\ast$$

Analogously, $b(a + b)^\ast \leq (a+b)^\ast$.

From both previous statements, again using monotonicity (ER-2):

$$ab(a+b)^\ast \leq a(a+b)^\ast \leq (a+b)^\ast.$$

To summarize: so far, we know that $a$,$b$ and $ab$ are all $\leq (a+b)^\ast$.

Now, observe that from (ER-1), it follows that if $x \leq w$ and $y \leq w$, then $x + y \leq w + y \leq w + w = w$. Thus, from $a\leq (a + b)^\ast$, $ab \leq (a + b)^\ast$ and $b \leq (a + b)^\ast$, you obtain:

$$(a + ab + b)(a+b)^\ast = a(a+b)^\ast + ab(a+b)^\ast + b(a+b)^\ast \leq (a + b)^\ast.$$

Since additionally $1 \leq (a+b)^\ast$ (directly from Ax-1), you obtain:

$$1 + (a + ab + b)(a+b)^\ast \leq (a+b)^\ast.$$

With (Ax-3) it follows:

$$(a + ab + b)^\ast = 1\cdot(a + ab + b)^\ast \leq (a + b)^\ast$$



Thus, both inequalities hold, hence we get the equality.

You cannot simply "remove" $^\ast$, nothing is told about "inverse" in the definition of a Kleene algebra. I also don't know what you mean by "$\ast$-height".