Your guess would have been true if $dW$ were just an ordinary differential. But this is not the case. $dW$ behaves somewhat different from the ordinary differential, in that intuitively we have $dW \approx \sqrt{dt}$ unlike to ordinary case. That's why the Ito,s formula is so important.
Now let us make a heuristic calculation. Let $Z_t = f(S_t) = \log S_t$, where $f(x) = \log x$. Then by the Ito's formula, we have
Z_t - Z_0 = f(S_t) - f(S_0) = \int_{0}^{t} f'(S_s) \; dS_s + \int_{0}^{t} \frac{1}{2} f''(S_s) \; dS_s^2,
where $dS^2$ is a formal symbol given by the product rule
$dW_s^2 = ds \quad \text{and} \quad dW_s ds = ds dW_s = 0 = ds^2.$
Thus $dS_s^2 = \sigma^2 S_s^2 dW_s^2$, and hence
$ \begin{align*} Z_t - Z_0 & = \int_{0}^{t} \frac{1}{S_s} \; (\mu S_s ds + \sigma S_s dW_s) + \int_{0}^{t} \left( -\frac{1}{2S_s^2}\right) \; (\sigma^2 S_s^2 ds) \\ & = \left( \mu - \frac{\sigma^2}{2}\right) t + \sigma W_t. \end{align*}$
Plugging $Z_t = \log S_t$ back and exponentiating, we obtain
$ S_t = S_0 \exp \left[ \left( \mu - \tfrac{\sigma^2}{2}\right) t + \sigma W_t \right]$