Since $b\le c$, then there is a $d$ such that $c=b+d$. Then
Note that this proof does not use the axiom of choice.
If you assume the axiom of choice, then $c+c=c$ for all infinite cardinals $c$. Without choice, it is consistent that this fails in general (see the comments here), though of course it holds in particular cases (for any well-ordered $c$, or for $c=\mathfrak c:=|\mathbb R|$, for example).
That $c\cdot c=c$ for all infinite $c$ is equivalent to choice. To prove that choice gives us this, it suffices to argue that $c\cdot c=c$ whenever $c$ is well-orderable. Typically, one uses some kind of pairing function to achieve this. See for example here. The point is that pairing functions give us explicit bijections between $\kappa\times \kappa$ and $\kappa$ for any infinite well-ordered cardinal $\kappa$. Since $\kappa\le\kappa+\kappa\le \kappa\cdot \kappa$, it follows from this that $\kappa+\kappa=\kappa$ as well.
To see that $c\cdot c=c$ for all infinite $c$ gives us choice, see here.
And yes, if $a+b=b$, then $a\le b$. On the other hand, without assuming choice, just knowing that $a\le b$ does not suffice to conclude that $a+b=b$. For example, if $b$ is infinite and Dedekind finite, then $1<b<b+1$.