There are several questions here, and I will try to answer them in order.
First, an example of an algebra which is not basic would be the $k$-algebra $M_2(k)$ of $2\times 2$ matrices over a field $k$. To see that it is not basic, note that as a right module over itself, it is the direct sum of the two submodules $R_1$ and $R_2$ of matrices with zero entries in row $2$ and row $1$, respectively. Since $R_1$ and $R_2$ are isomorphic as right $M_2(k)$-modules, the algebra is not basic.
Second, on the existence of certain sets of idempotents. I am assuming here that you are interested in complete sets of pairwise orthogonal primitive idempotents, that is, collections of elements $e_1, \ldots, e_n$ such that $e_i\cdot e_j$ is $e_i$ if $i=j$ (idempotence) and $0$ otherwise (orthogonality); $\sum_{i=1}^n e_i = 1$ (completeness); and none of the $e_i$ can be written as a sum of two orthogonal non-zero idempotents (primitiveness). If an algebra is artinian, then such a set of idempotents always exists. An algebra which does not admit such a set is $\prod_{\mathbb{N}} k$, the product of a countably infinite number of copies of a field $k$. This algebra admits complete sets of pairwise orthogonal idempotents, but we cannot require that these idempotents are all primitive. Another example would be the $\mathbb{R}$-algebra of continuous functions from $\mathbb{R}$ to $\mathbb{R}$.
Next, you ask about the quiver of a non-basic algebra. This is somewhat ill-defined. Let's review the theory over an algebraically closed field $k$. In this setting, for any basic, finite-dimensional $k$-algebra $A$, there exists a unique quiver $Q$ and a (non-unique) admissible ideal $I$ of $kQ$ such that $A$ is isomorphic to $kQ/I$. Moreover, any algebra of the form $kQ/I$ as above is basic. It follows from these results that if $A$ is not basic, then it cannot be of the form $kQ/I$, with $I$ an admissible ideal of $kQ$.
In the above discussion, the operative word is admissible. For any finite-dimensional $k$-algebra, you can find lots of quivers $Q$ and (not necessarily admissible) ideals $I$ such that $A$ is isomorphic to $kQ/I$. For matrix algebras $M_n(k)$, some such descriptions are given in this mathoverflow question and its answers. Still, for non-basic algebras, it will never be possible to find an admissible ideal, and there will be no good notion of the uniqueness of the quiver.
Finally, if an algebra is basic but not connected, then the description by a quiver with an admissible ideal of relations works fine; all that happens is that the quiver is not connected.
I'll assume that the quiver $Q$ has finitely many vertices and arrows.
Then even if $Q$ has oriented cycles, it is it is still true that the path algebra $\mathbb{C}Q$ determines the quiver $Q$.
(1) In the case of a quiver with no oriented cycles, a quick way to recover the quiver is as follows:
There is a simple module $S_i$ associated with each vertex $i$, and the number of arrows from vertex $i$ to vertex $j$ is $\dim_{\mathbb{C}}\operatorname{Ext}^1_{\mathbb{C}Q}(S_i,S_j)$. [I'm not sure that this is the proof that Derksen was asking for, as the book doesn't seem to assume knowledge of $\operatorname{Ext}$.]
So knowing the simple $\mathbb{C}Q$-modules and the extensions between them lets you recover the quiver.
(2) If the quiver has oriented cycles but no loops (arrows with the target equal to source) then there are more simple modules, but the obvious simples associated to the vertices are the only $1$-dimensional simples, and the same method of recovering the quiver works, if you only consider the $1$-dimensional simples.
(3) If the quiver has loops, then there are more $1$-dimensional simples (consider the representation with $\mathbb{C}$ at vertex $i$, zero at every other vertex, with the loops at vertex $i$ acting by multiplication by arbitrary scalars). But the same method work if we can pick out one $1$-dimensional simple module for each vertex.
There may be a simpler method, but one way to do this is to consider the abelianization of $\mathbb{C}Q$. This is a product of polynomial algebras $\mathbb{C}[x_1,\dots,x_{r_i}]$, one for each vertex $i$, where $r_i$ is the number of loops at vertex $i$. So it has one primitive idempotent for each vertex, and for each of these idempotents we can choose any $1$-dimensional simple module (it doesn't matter which) that is not annihilated by that idempotent. As before, the quiver can then be recovered by considering $\operatorname{Ext}^1_{\mathbb{C}Q}$ between these simple modules.
Best Answer
In the finite-dimensional world over algebraically-closed fields the path algebras are in fact only the hereditary algebras (i.e. the algebras such that $\operatorname{Ext}^2$ vanishes) up to Morita equivalence.