For example $f_n(x)=(1+x/n)^n$ converges pointwise on $\Bbb R$ to $f(x)=e^x$, but not uniformly because $f_n(n)\to+\infty$ and $f_n(-2n)$ has no limit.
Is it logically sound to say that the convergence is uniform on every compact $E_M=[-M,M]$ because $\lvert f_n(x)-f(x)\rvert$ is continuous on $E_M$ for all $n$, and thus attains a maximum $\left\lvert\left(1+\frac{a_n}{n}\right)^n-e^{a_n}\right\rvert$ which goes to $0$ as $n\to\infty$ because $a_n$ is confined in $[-M,M]$?
Your proof is not good. Take, for example
$$f_n(x) = \begin{cases}nx & 0\leq x\leq \frac{1}{n}\\ 2-nx & \frac{1}{n}\leq x\leq \frac2n\\ 0& x\geq \frac2n\\ \end{cases}$$
Clearly, $f_n-f$ is a continuous function on $[0,1]$ for every $n$. It attains a maximum at $a_n$, and the maximum is confined in $[0,1]$.
However, we cannot conclude that convergence is uniform on $[0,1]$, because it is not.
The issue with your proof is in saying that the maximum "goes to zero" without actually proving that it goes to zero. You claim that it's because $a_n$ is confined on a compact interval, but as you see from my proof, that's not enough.