Given K a field, and $a,b \in K, a\ne0$, show $ax+b=0$ has only one solution.
is it possible to just say $ax=-b \rightarrow x=a^{-1}*-b$, we can multiply by $a^{-1}$ because each element has inverse.
Does this prove the linear equation has only one solution?
On
If you use $\iff$ as much as possible in proofs you will more likely avoid the mistake of only proving half of what is needed. For example:
If $a\ne 0$ then $a^{-1}$ exists; therefore $$ax+b=0\iff a^{-1}(ax+b)=a^{-1}0=0 \iff$$ $$\iff x+a^{-1}b=0\iff$$ $$\iff (x+a^{-1}b )+(-a^{-1}b)=0+(-a^{-1}b)=-a^{-1}b\iff$$ $$\iff x=-a^{-1}b.$$ BTW. The justification for the first "$\iff$" , above, is that $$ax+b=0\implies a^{-1}(ax+b)=a^{-1}(0)\implies$$ $$\implies a(a^{-1}(ax+b))=a(a^{-1})(0)\implies$$ $$\implies (a\cdot a^{-1})(ax+b)=(a\cdot a^{-1})(0)\implies$$ $$\implies (1)(ax+b)=(1)(0)\implies$$ $$\implies ax+b=0.$$ That is , when you have a string such as $A\implies B\implies C\implies D \implies A,$ then you have $A\iff B.$
That's ok; but you only used one-directional implication, sp now you need to go back and check that that is a solution indeed. Without checking, what you showed is that"If the equation has a solution, then it is $-b a^{-1}$. So it's possible it has that solution, and it's also possible it has no solutions, if you don't check that that's always a solution.
Alternatively you could just replace the one-directional arrow by a bi-directional arrow and you're done.