Equations for internal energy and their relevance to the derivation of the equation for entropy for perfect gases
Your equation $dU = TdS - pdV$ is valid for all pure substances, not just ideal gases.  It is telling you that if you knew an equation that explicitly defined $U$ as a function of $S$ and $T$, then you could calculate all other thermodynamic properties of that substance, i.e. if you knew that $U=f(S, V)$, then you could easily calculate $T$ and $p$ because $T=\frac{\partial f(S, V)}{\partial S}$ and $p=\frac{\partial f(S, V)}{\partial V}$.  This is because $S$ and $V$ are canonical variables for $U$.

Now let's apply the ideal gas equation $dU=C_V dT$.  You can integrate this to yield $U=C_V T + K$.  Unfortunately, this does not express $U$ as a function of $S$, but rather of $T$.  So how do we proceed?  This physics.se question has a good general answer.

But in this specific case things are bit easier.  Let's find $dS$ from the equation you were given.

$dS =\frac{1}{T}(dU + p dV)=\frac{1}{T}(C_V dT + \frac{RT}{V}dV)=C_V \frac{dT}{T}  + R \frac{dV}{V}$

Since you now know $dS$ and from the ideal gas law you also know $p dV$, you have an expression for $dU$ from which you can derive all thermodynamic quantities.