I just wanted to add something to the discussion about the utility of adding additional basepoints. It turns out this is crucial for understanding certain aspects of embedding theory. See the bottom of this answer for an explanation.
For a map of spaces $A \to B$, let $\text{Top}(A\to B)$ be the category of spaces which factorize this map. This has objects given by factorizations $A \to X \to B$ and morphisms maps $X \to X'$ which are factorization compatible in the obvious sense. Let's consider the case of the constant map $S^0 \to \ast$. Clearly, an object of $\text{Top}(S^0\to \ast)$ is just a space with a preferred pair of basepoints.
Then unreduced suspension can be regarded as a functor $$ S: \text{Top}(\emptyset \to \ast) \to \text{Top}(S^0 \to *) . $$ That is, the functor which assigns to an unbased space its unreduced suspension, considered as a space with two basepoints.
Now a desuspension question in this context asks given an object $X \in \text{Top}(S^0 \to *)$, is there an object $Y \in \text{Top}(\emptyset \to \ast) $ and a weak equivalence $$ SY \simeq X ? $$ More generally, I've gotten a lot of mileage out of the fiberwise version of this question.
Given a space $B$ we can consider the unreduced fiberwise suspension of $\emptyset \to B$ as the projection map $B \times S^0 \to B$ (here unreduced fiberwise suspension of $Y\to B$ means the double mapping cylinder of the diagram $B \leftarrow Y \to B$, or concretely, it's $B \times 0 \cup Y \times [0,1]\cup B \times 1$.
Unreduced fiberwise suspension is then a functor $$ S_B: \text{Top}(\emptyset \to B) \to \text{Top}(B\times S^0 \to B) , $$ and one can consider the problem of whether an object $X \in \text{Top}(B\times S^0 \to B)$ can be written as $S_B Y$ up to weak equivalence.
Why I care about this problem
This problem naturally arises in embedding theory: if $P \to N \times [0,1]$ is an embedding, where $P$ and $N$ are closed manifolds and if $W$ is the complement of $P$ in $N \times [0,1]$ then $W$ is an object of the category $\text{Top}(N\times S^0 \to N$) and a necessary obstruction to compressing $P$ as an embedding into $N$ is that $W$ should fiberwise desuspend over $N$. Furthermore, in certain instances the existence of fiberwise unreduced desuspension suffices to finding the compression of the embedding. (This story is explained in detail in the paper: Poincaré duality embeddings and fiberwise homotopy theory, Topology 38, 597–620 (1999).)
Postscript
In the fiberwise context there is a real difference between the reduced and unreduced cases of the desuspension problem. For example, in the case of the compression problem $P \to N \times I$ described above, the two inclusions $N \times i \to P$ for $i = 0,1$ might have distinct (fiberwise) homotopy classes. If this is the case, then there's no chance that the complement data $W$ can underly a reduced fiberwise suspension, for if it did, then the map $N \times S^0 \to P$ would factor through $S_N N \cong N \times D^1$, giving a homotopy of the two inclusions $N \times i \to P$.
(For $Y \in \text{Top}(\text{id}:B \to B)$, the reduced fiberwise suspension $\Sigma_B Y$ is given by $$ \Sigma_B Y = \text{colim}(B \leftarrow S_B B \to S_B Y) . $$ This is an endo-functor of $\text{Top}(\text{id}:B \to B)$.
An even more mundane example is this: when $B = \ast$, we can consider $S^0$ with its two distinct basepoints. Clearly $S^0 = S\emptyset$, but $S^0$ is not, even up to weak equivalence, the reduced suspension of any based space.
