I am making no claims about the physics here, just a way to clarify the mathematics presented in the question. More/different work may need to be done to bring this in line with the standard physical interpretations of the equations.
1.
Suppose, as an example, that $E(V,T,O)$ and $S(V,T,O)$ are some sort of functions on real variables: $E$ and $S$ both depending on $V$, $T$, and some other factor $O$. Now, for some physical reason, we might be only considering these functions with one or more constraint equations, like maybe $T+O=7$.
In any case, $\dfrac{\partial E}{\partial V}$ would depend on the values of the other inputs $T,O$. If we consider $\dfrac{\partial E}{\partial V}$ at the particular values $T=T_0$ and $O=O_0$ (perhaps from context), then $\dfrac{\partial E}{\partial V}=\dfrac{\mathrm d}{\mathrm dV}E(V,T_0,O_0)$. So a partial derivative is a total derivative with $T$ (and any other things that aren't $V$) held constant.
If we assume $T=T_0$ (and everything else constant), then the equation with the partial derivatives becomes $\dfrac{\mathrm dE}{\mathrm dV}=T_0\dfrac{\mathrm dS}{\mathrm dV}-P$. In this context with everything else held constant, we could solve for $P$ as a function of $V$: ($P=T_0\dfrac{\mathrm dS}{\mathrm dV}-\dfrac{\mathrm dE}{\mathrm dV}$), so it makes sense to speak of $\dfrac{\mathrm dP}{\mathrm dV}$.
One way to get to the equation with the $\mathrm d$s is to "multiply through by $\mathrm dV$" without worrying about definitions. But another perspective is through "differential forms" where, by definition, if $f(V)$ is a function of a single variable, we have $\mathrm df=\left(\dfrac{\mathrm df}{\mathrm dV}\right)\mathrm dV$ (where $dV$ is just an odd-looking other input variable for $\mathrm df$). Then multiplying both sides by $\mathrm dV$ does give you $\mathrm dE=T_0\mathrm dS-P\mathrm dV$ by definition (here $\mathrm dV$ is like an independent variable).
2.
It's not common to take partial derivatives of infinitesimals. But if you're working with differential forms, and $f(x,y,z)$ is a function, then $\mathrm df=\dfrac{\partial f}{\partial x}\mathrm dx+\dfrac{\partial f}{\partial y}\mathrm dy+\dfrac{\partial f}{\partial z}\mathrm dz$, for example.
3.
Depending on what you want to know about differential forms, there's a lot of stuff to read, but you may want to start with this MSE question about how a derivative is/is not a ratio. It's also possible that things are complicated by physicists using notation slightly differently to how mathematicians would commonly write things.
