The question is in the title. The solution gives this: "No; the set of isomorphisms does not contain the zero map (unless the space is trivial)."
To be honest, I have no idea how to solve this. I define subspace to be a subset, and my initial thoughts here is that a mapping between a vector space V to itself that's also one to one and onto is just a subset of mappings between a vector space V to itself that's not necessarily one to one or onto. How exactly does the zero map come into play here? Why does the zero map need to be in the set of isomorphisms in order for them to be considered a subspace of the space L(V, V)?
That's incorrect. A subspace of a vector space $V$ is a subset $W\subseteq V$ which is also a vector space in its own right (with the addition and scalar multiplication "coming from $V$" in the obvious way). This means that $W$ must be
closed under addition (for each $a,b\in W$ we also have $a+b\in W$),
closed under scalar multiplication (for each scalar $s$ and each $a\in W$ we also have $s \cdot a\in W$), and
contain the zero vector,
where "$+$," "$\cdot$," and "zero vector" all take their meaning from $V$.
This more nuanced meaning of "sub[thing]" is ubiquitous in mathematics; compare "subgroup," "subring," ...
For example, consider $\mathbb{R}^2$ as a vector space over $\mathbb{R}$ in the usual way. The subset $$\{(0,r):r\in\mathbb{R}\}$$ is a subspace of $\mathbb{R}^2$, but $\{(1,r): r\in\mathbb{R}\}$ is not (in fact the latter fails all three of the bulletpoints above). The set $L(V,V)$ of your question is more difficult to picture, but the point is the same: the "zero vector" of $L(V,V)$ is the constant map $${\bf v}\mapsto{\bf 0}_V,$$ but this is usually not an isomorphism from $V$ to itself.
Exercise: when is the zero map an isomorphism?