Exactly right, your friend used a function that is not a one to one function, and for those functions, we cannot claim $f(a)=f(b)\implies a=b$. In order to do that, your friend would first have to demonstrate that the function he is using is one to one, but what he in fact did was show that it is not one to one, since he discovered two values $a,b$ such that $a\neq b$ and $f(a)=f(b)$.
So, what your friend thought he did:
Take this one-to-one function, and voila, I have shown that $1=0$
What he really did:
Take this function, and assume that it is one-to-one. Then it follows that $1=0$.
and this is just one step away from a proof by contradiction that the function is actually not one-to-one.
The actual case is:
$a^x$ is not strictly increasing for $a>0$. It is strictly increasing for $a>1$, and is constant for $a=1$ (and actually strictly decreasing for $0<a<1$!)