First, your argument is flawed. I assume the theorem you are referring to is that you have uniform convergence on $[−a,a]$ for any real number $a$ with $a<R$, where $R$ is the radius of convergence of the power series.
But you can not take $a=\infty$ in your problem...
What you could use is the following, very useful, fact:
A series $\sum\limits_{k=1}^\infty f_k(x)$ converges uniformly on $I$
if and only if the sequence of partial sums $S_n=\sum\limits_{m=1}^n
f_m$ is uniformly Cauchy. That is, for every $\epsilon>0$,
there is an integer $N$ such that
$$|S_m(x)-S_n(x)|=\Bigl|\sum_{k=n+1}^m f_k(x)\Bigr|<\epsilon,\quad\text{ for all }x\in I\ \ \text{and all }m\ge n \ge N.$$
An immediate result of this (and a result that, by itself, is easily proven, as in Davide Giraudo's answer) is:
If $\sum\limits_{k=1}^\infty f_k(x)$ converges uniformly on $I$, then
for every $\epsilon>0$, there is an integer $N$ such that $$|f_n(x)|
=| S_{n }(x)-S_{n-1}(x)|<\epsilon,\quad\text{ for all }x\in I\ \ \text{ and all }n \ge N;$$ that is, the sequence $(f_n)$ converges uniformly
to the zero function on $I$.
One can use the first result I mentioned directly to show that your series is not uniformly Cauchy over $\Bbb R$:
Since $a_i\ne 0$ for infinitely many $n$, given any positive integer $N$, you can choose $m>n\ge N$ such that$\sum\limits_{k=n}^m a_k x^k$ is not a constant polynomial. Taking the limit as $x$ tends to infinity will verify that this sum is not uniformly close to 0.
