"Weaker" postulates for a group? In some learning resources I have, it has been claimed that the group axiom defining the identiy $e$ $$a \cdot e = e \cdot a = a$$ and the group axiom defining the inverse element $$a \cdot b = b \cdot a = e$$ can be replaced by the "weaker" statements $$a \cdot e = a$$ and $$a \cdot b = e$$.
The argument proceeds by asking the reader to consider elements $a$, $b$ and $c$ in group $G$ such that $a \cdot b = e$ and $b \cdot c = e$, then $$b \cdot a = (b \cdot a) \cdot e = (b \cdot a) \cdot (b \cdot c) = b \cdot (a \cdot b) \cdot c = b \cdot e \cdot c = b \cdot c = e$$
I am unhappy with this, because at the very end, the claim that $b \cdot e \cdot c = b \cdot c$ relies on the fact that $e \cdot c = c$, but we are supposed to only be working from $c \cdot e = c$
The other weaker form is then "proved" with $$a = a \cdot e = a \cdot (b \cdot a) = (a \cdot b) \cdot a = e \cdot a$$ which is all very well except for the fact that it now uses $a \cdot b = b \cdot a = e$ which I feel has been proved on shaky ground.
Am I right to be concerned with this argument/approach, or are my logic circuits shortcircuiting somewhere?
Is the assertion that the properties/axioms/postulates for a group can be replaced with such a "weaker" form correct?
 A: It is correct that it suffices to expect the presence of a right identity and a right inverse for every element. It is more direct to immediately deduce that given the other axioms (well, using associativity) every right identity will be a left identity and every right inverse will be a left inverse. All of this holds too if you swap the role of right and left.
This relies on the following: $a^2=a$ implies that $a=e$. This fact is already deducible given only a one-sided identity and one-sided inverse (but it has to be the same side! I advise you to come up with a proof). Indeed, suppose that $ab=e$ then
$$
(ba)^2=(ba)(ba)=b(ab)a=b(e)a=(be)a=ba\implies ba=e
$$
from which we then have
$$
ea=(ab)a=a(ba)=ae=a\,.
$$

Regarding your concerns:

the claim that $b⋅e⋅c=b⋅c$ relies on the fact that $e⋅c=c$, but we are supposed to only be working from $c⋅e=c$

We are only using $ce=c$! Indeed, $bec=(be)c=b(ec)$. While we know nothing about $ec$ per se we know something about $be$ and associativtiy allows us to  alternate between these two expressions. This was already pointed out in the comments by mxian.
