This is my attempt at a proof:
$\forall x \in \mathbb{R}, \lim_{n \to \infty}(1+\frac{x}{n})^n=\lim_{m \to 0}\frac{log(1+mx)}{m}=\lim_{m \to 0} \frac{x}{1+mx}=x \iff (1+\frac{x}{n})^n \rightarrow e^x$ uniformly.
It follows that $(1+\frac{x}{n})^n e^{-x} \rightarrow 1$ uniformly.
Is this sufficient?
Let $f_n(x) = \left(1+\frac xn \right)^n e^{-x}$. In order for $f_n$ to converge to $f\equiv 1$ uniformly, we must have $$\lim_{n\to\infty}\sup_{x\in\mathbb R}|f_n(x)-1|=0. $$ Since $f_n\in\mathcal C^1(\mathbb R)$ and $\mathbb R$ is not bounded, any minimum or maximum value must be obtained as a critical point, i.e. a point $x$ such that $f_n'(x)=0$. Computing the derivative: $$ f_n'(x) = -\frac xn\left(1+\frac xn\right)^{n-1}e^{-x}, $$ it follows that $f_n'(x)=0$ implies $x=0$ or $x=-n$. As $f_n(-n)=0$, it follows that $$\lim_{n\to\infty}\sup_{x\in\mathbb R} |f_n(x)-1| \geqslant 1, $$ and hence $f_n$ does not converge uniformly.
Edit: if $f_n$ is defined on $[0,1]$ instead of $\mathbb R$, then $n\notin[0,1]$ for $n>1$, so a minimum/maximum value must be obtained on the boundary of $[0,1]$. We compute $f_n(0) = 1$ and $$f_n(1) = \left(1+\frac1n\right)^n e^{-1}\stackrel{n\to\infty}\longrightarrow 1, $$ so indeed $f_n$ converges uniformly on $[0,1]$.
Edit: if $f_n$ is defined on a compact set $K\subset\mathbb R$, then $K\subset[-M,M]$ for some $M>0$. Then choosing $n>M$, we see that again an extreme value of $f_n$ must be obtained at $x=M$ or $x=-M$. Since $$f_n(-M) = \left(1 - \frac Mn\right)^n e^M\stackrel{n\to\infty}\longrightarrow 1$$ and $$f_n(M) = \left(1 + \frac Mn\right)^n e^{-M}\stackrel{n\to\infty}\longrightarrow 1,$$ we see again that $f_n$ converges uniformly to $1$.