There are many different things going on here. One is that to invert the suspension functor is to stabilize. The Freudenthal suspension theorem tells you that the system of suspension maps $[X,Y] \to [SX,SY] \to [S^2X,S^2Y] \to \cdots$ eventually stabilizes (at least for finite CW-complexes), and so you can view this as a simplification of the usual category of spaces. Moreover, for formal reasons this implies that your category is enriched in abelian groups, which gives you more traction on things. (For instance, you can always add two maps, maps always have inverses, etc.)
More broadly, the real power of homotopy theory is that pretty much everything is representable in one way or another, which means that you can turn your tools on each other and get them to tell you new things about the tools themselves. From this point of view, it's entirely natural to define the category of spectra (since it is equivalent to the category of cohomology theories). But it sounds like you already knew about this piece of motivation.
This is a general fact about sequential colimits (or even more generally colimits over posets). $\newcommand\colim{\operatorname{colim}}$
Let $A : \mathbb{N}\to \mathcal{C}$ be some sequential diagram.
$\mathbb{N}$ has an endomorphism $s$ defined by $s(n)=n+1$.
There is a corresponding natural transformation $\sigma : A \to s^*A$,
which at $n$ is the map $f_n : A_n\to A_{n+1}$ which is $A(n\le n+1)$, the transition map in the sequential diagram.
In fact, more generally, let $t : \mathbb{N}\to \mathbb{N}$ be any nondecreasing map, then $t$ gives an endomorphism of $\mathbb{N}$ as a category, so we can consider the pullback $t^*A$. If $t$ additionally satisfies $t(n)\ge n$ for all $n$, then there is a natural transformation $\tau: A\to t^*A$ which at $n$ is the unique map $A_n\to A_{t(n)}$ in the diagram $A$.
Now if we assume that $t$ is cofinal (for all $n\in \mathbb{N}$, there exists $N\in\mathbb{N}$ such that $t(N)\ge n$), the following is true.
Proposition. $\colim A$ exists if and only if $t^*\colim A$ exists. In this case if $t(n)\ge n$ for all $n$, so $\tau$ exists, then $\tau$ induces a morphism $\tilde{\tau}:\colim A\to \colim t^* A$ on passing to colimits which is an isomorphism.
Proof.
It suffices to give a natural isomorphism between $\newcommand\cocone{\operatorname{co-cone}}\cocone_A$ and $\cocone_{t^*A}$, the functors that the colimits represent. Given some cocone to $A$, $(X, i_n: A_n\to X)$, we get a cocone to $t^*A$ by taking $(X, i_{t(n)}: A_{t(n)}\to X)$. However, it turns out that we can also go backwards, since $t$ is cofinal. If we are given $(Y,j_m : A_{t(m)}\to Y)$, we can define
$i_n : A_n\to Y$ by taking $N$ large enough that $t(N)\ge n$, and then defining $i_n$ to be the composite $A_n \to A_{t(N)} \xrightarrow{j_N} Y$. This is well defined regardless of the choice of $N$ because $\mathbb{N}$ is a poset and $(Y,j_m)$ are a cocone to $t^* A$. This gives us a cocone $(Y, i_n)$ to $A$. These operations are inverse to each other, so we have a natural isomorphism as desired.
Finally we need to show that when $t(n)\ge n$ for all $n$, and if $\colim A$ exists then $\tau$ induces an isomorphism on the colimits. First let's fix some notation. Let $i_n : A_n\to \colim A$ be the inclusions for the colimit of $A$, let $j_n : A_{t(n)}\to \colim t^*A$ be the inclusions of the colimit of $t^*A$, and let $\tau_n: A_n\to A_{t(n)}$ be the components of $\tau$.
Then we observe that $\tilde{\tau}$ is defined by
applying the universal property of the colimit $\colim A$ to the cocone
$(\colim t^*A, j_n \circ \tau_n)$. But we know (up to isomorphism) what the $j_n$ are. By the first half of this proof the $j_n$ are $i_{t(n)}$, so
$$j_n\circ \tau_n = i_{t(n)}\circ \tau_n = i_n,$$
since $\tau_n : A_n\to A_{t(n)}$ is a map in the diagram $A$ and $i$ is a
cocone to $A$. Thus after potentially composing with an isomorphism $(\colim t^*A, j_n)\to (\colim A, i_{t(n)})$, $\tilde{\tau}$ is actually given by the colimiting cone itself, so we have that for some isomorphism $\phi$, $\phi\circ \tilde{\tau} = 1_{\colim A}$, and thus $\tilde{\tau}$ is an isomorphism itself (namely $\phi^{-1}$). $\blacksquare$
Edited to fix the final argument
Build a modified version of the stable homotopy group colimit sequence as follows.
Fix $a$, let $P_{2n} = \pi_{n+a}(X_a)$, let $P_{2n+1} = \pi_{n+a}(\Omega\Sigma X_a)$, let the maps $f_{2n} : P_{2n}\to P_{2n+1}$ be $\pi_{n+a}(\eta_{X_a})$, and let the maps $f_{2n+1} : P_{2n+1}\to P_{2n+2}$ be given by the composite of $\pi_{1+n+a}$ applied to the structure map $\Sigma X_a\to X_{a+1}$ with the isomorphism $\pi_{n+a}(\Omega\Sigma X_a)\cong \pi_{n+1+a}(\Sigma X_a)$.
Let $d: \mathbb{N}\to \mathbb{N}$ be given by $d(n) =2n$, and let
$\delta : P\to d^*P$ be the corresponding natural transformation.
Note that $d^*P$ is precisely the sequence that usually defines the stable homotopy groups.
Now consider the $\sigma$ natural transformation, $\sigma_P$, for this sequence. If we restrict to even $n$, then $\sigma_P$ corresponds to $\pi_{\bullet+a}(\eta)$. In other words $d^*\sigma_P = \pi_{\bullet + a}(\eta)$. We have a commutative diagram
$$
\require{AMScd}
\begin{CD}
P @>\sigma_P>> s^* P\\
@V\delta VV @V\delta VV \\
d^*P @>>\pi_{\bullet+a}(\eta)> d^*s^*P.\\
\end{CD}
$$
Termwise this commutative diagram is
$$
\require{AMScd}
\begin{CD}
P_n @>>> P_{n+1}\\
@VVV @VVV \\
P_{2n}=\pi_{n+a}(X_a) @>>\pi_{n+a}(\eta_{X_a})> P_{2n+1}=\pi_{n+a}(\Omega \Sigma X_a).\\
\end{CD}
$$
Now if we take colimits of this diagram of functors, three of the maps, $\sigma_P$ and both $\delta$s induce isomorphisms on the colimit. Therefore the last map does as well.
Best Answer
Let $\mathscr{C}$ be any $\infty$-category with finite colimits and a terminal object $*$. We denote $\mathscr{C}_*=*/\mathscr{C}$. Let $\mathsf{A}$ be the diagram shape $2\leftarrow 0\to 1$, and let $i_0\colon[0]\to\mathsf{A}, 0\mapsto 0$ be the inclusion of the initial vertex, with corresponding right Kan extension $\mathrm{Ran}_{i_0}\colon\mathscr{C}\to\mathrm{Fun}(\mathsf{A},\mathscr{C})$ (informally sending $X$ to a diagram $*\leftarrow X\to *$). Let $\mathsf{Sq}$ be the commutative square category $$ \require{AMScd} \begin{CD} 0 @>>> 1\\ @VVV @VVV\\ 2@>>> 3 \end{CD} $$ with corresponding inclusion functor $i\colon\mathsf{A}\to\mathsf{Sq}$. This gives us a corresponding left Kan extension $\mathrm{Lan}_{i}\colon\mathrm{Fun}(\mathsf{A},\mathscr{C})\to\mathrm{Fun}(\mathsf{Sq},\mathscr{C})$, which takes a colimit of the span to ''complete the square''. Let $\mathsf{E}$ be the one-arrow category $1\to 3$, with corresponding inclusion functor $j\colon\mathsf{E}\to\mathsf{Sq}$. We define a functor $\widetilde{\Sigma}_+\colon\mathscr{C}\to\mathscr{C}_*$ as the composite $$ \mathscr{C}\xrightarrow{\mathrm{Ran}_{i_0}}\mathrm{Fun}(\mathsf{A},\mathscr{C})\xrightarrow{\mathrm{Lan}_{i}}\mathrm{Fun}(\mathsf{Sq},\mathscr{C})\xrightarrow{j^*}\mathrm{Fun}(\mathsf{E},\mathscr{C})\simeq\mathrm{Ar}(\mathscr{C}), $$ which lands in the full subcategory of the arrow category $\mathrm{Ar}(\mathscr{C})$ on the arrows with $*$ as domain. Hence $\widetilde{\Sigma}_+$ factors through the inclusion $\mathscr{C}_*\to\mathrm{Ar}(\mathscr{C})$. Denote by $\Sigma_+\colon\mathscr{C}\to\mathscr{C}_*$ the resulting functor. Writing $U\colon\mathscr{C}_*\to\mathscr{C}$ for the forgetful functor, we of course have $\Sigma\colon\mathscr{C}\to\mathscr{C}$ to be naturally equivalent to $U\Sigma_+$. In a similar way, we can formally define the suspension functor $\mathrm{Sigma}_*\colon\mathscr{C}_*\to\mathscr{C}_*$. Given an object $(X,x)$ of $\mathscr{C}_*$, the object $\mathrm{Sigma}_*(X,x)$ is naturally equivalent to $\Sigma X$ equipped with the basepoint $*\xrightarrow{x}X\to\Sigma X$. However, since $X\to\Sigma X$ factors as $X\to *\to \Sigma X$, where $*$ is the terminal object at position $1$ in $\mathsf{A}$ and $\mathsf{Sq}$, we find that this basepoint is naturally equivalent to the basepoint that $\Sigma X$ receives from its lift $\Sigma_+X$. In fact, we find that $\Sigma_*(X,x)\simeq\Sigma_+U(X,x)$ naturally in $(X,x)$.
This means that the diagram
of $\infty$-categories and functors commutes. This in turn implies that
$$
\mathrm{colim}(\mathscr{C}\xrightarrow{\Sigma}\mathscr{C}\xrightarrow{\Sigma}\ldots)\simeq\mathrm{colim}(\mathscr{C}_*\xrightarrow{\Sigma_*}\mathscr{C}_*\xrightarrow{\Sigma_*}\ldots),
$$
so the two definitions agree. (Note that, for general $\mathscr{C}$, neither of these $\infty$-categories will model spectrum objects in $\mathscr{C}$.)
That leaves the question why we prefer to work with the pointed version. I can think of a couple of reasons: firstly, pointed categories are easier to work with than unpointed ones. Having a zero map can for instance make it easier to make certain constructions natural. Secondly, the suspension functor $\Sigma$ as defined above is generally not a left adjoint even if $\mathscr{C}$ has finite limits, while $\Sigma_*$ is in that case (from now on, just assume $\mathscr{C}$ has finite limits). $\Sigma_+$ is also not a left adjoint: you really need to choose both natural basepoints of $\Sigma X$ as such and consider the lift $\Sigma_{++}\colon\mathscr{C}\to\mathscr{C}_{*\sqcup*/}$ to get a right adjoint functor, namely the loop functor $\Omega$. But as said, $\Sigma_*$ is just immediately a left adjoint. This is particularly helpful if you don't work $\infty$-categorically: if you work with model categories, then your main way to model $\infty$-functors is by building Quillen adjunctions. In particular, you want a good supply of adjoint $1$-functors, so you prefer to work with the pointed version. But also $\infty$-categorically we prefer adjoint functors over non-adjoint ones. Thirdly, the equivalence $$ \mathrm{colim}(\mathscr{C}\xrightarrow{\Sigma}\mathscr{C}\xrightarrow{\Sigma}\ldots)\xrightarrow{\simeq}\mathrm{colim}(\mathscr{C}_*\xrightarrow{\Sigma_*}\mathscr{C}_*\xrightarrow{\Sigma_*}\ldots), $$ defined above requires a choice, namely the choice in the definition of $\Sigma_+$ to let one, and not the other natural candidate for the basepoint be the actual basepoint. You might object, and say that if we let $\Sigma_{-}$ denote the functor we would get by choosing the other basepoint, then the fact that both $\Sigma_+$ and $\Sigma_{-}$ induce on colimits inverses to the canonical equivalence $$ \mathrm{colim}(\mathscr{C}\xrightarrow{\Sigma}\mathscr{C}\xrightarrow{\Sigma}\ldots)\xleftarrow{\simeq}\mathrm{colim}(\mathscr{C}_*\xrightarrow{\Sigma_*}\mathscr{C}_*\xrightarrow{\Sigma_*}\ldots), $$ in the other direction (the one induced by $U$) would imply that both $\Sigma_+$ and $\Sigma_{-}$ induce up to equivalence the same map on colimits. And while that is true, it is not up to canonical equivalence. So there is a very slight non-canonical thing occurring because we had to choose whether the north or the south pole of our suspension was the base point. And while these can be connected by a path, the choice of path is not canonically determined up to higher homotopies. In the pointed setting, everything is completely canonical. There probably are more reasons (also of less philosophical and more practical nature), but I'll keep it at this for now.