Why isn't ΔG equal to zero for this equilibrium reaction? In the reaction $\ce{E(g) &lt;=&gt; P(g)}$ at 25 °C, equilibrium is reached when the pressure of the product is 0.100 that of the reactant. What is $\Delta G$ in joules?

Using $\Delta G = RT\ln K$, my answer is $-8.31 \times 298 \times \ln (1.00) \ \text{J}$.
However, the answer is $-8.31 \times 298 \times \ln (0.100) \ \text{J}$. Wouldn't this mean that $\Delta G &gt; 0$, therefore favoring the reactants? (As an aside, why would the reaction even be shifted to the left since it has a higher pressure?)
You are mixing up two different quantities. The equation you are using describes the standard Gibbs free energy of reaction $\Delta G^{0}$, i.e. $\Delta G$ for standard conditions without any requirement of the reaction being in equilibrium.
And this quantity can very well be different from zero.
For non-standard conditions the Gibbs free energy of reaction is

\begin{equation}
 \Delta G = \Delta G^{0} + RT \log K
\end{equation}

and as you stated correctly this quantity must (per definition) be equal to zero for equilibrium conditions.

For a derivation see this answer of mine.