This claim occurs in the proof of Proposition 4F.1 in the version from 2002 (section "Spectra and Homology Theories").
It is not true, because in general $\Sigma^i X \vee \Sigma^i Y$ has cells of dimension $ > 2i-1$.
However, Hatcher only wants to prove that
$(\ast)$ $\pi_{n+i}(\Sigma^i X \vee \Sigma^i Y) \approx \pi_{n+i}(\Sigma^i X \times \Sigma^i Y)$ for $n+i < 2i-1$.
But now Tyrone's comment applies.
By the way, the following statement is correct and suffices to prove $(\ast)$:
$\Sigma^i X \vee \Sigma^i Y$ and $\Sigma^i X \times \Sigma^i Y$ have the same $(2i-1)$-skeleton.
The cells of $ \Sigma^i X \times \Sigma^i Y$ have the form $e \times e'$ with cells $e$ of $\Sigma^i X$ and $e'$ of $\Sigma^i Y$. We have $\dim(e \times e') \le 2i-1$ only in case $e = \{ \ast \}$ and $\dim(e') \le 2i-1$ or $e' = \{ \ast \}$ and $\dim(e) \le 2i-1$ because $\Sigma^i Z$ has one $0$-cell $\{ \ast \}$ (the basepoint) and all other cells have dimension $\ge i$.
$\Sigma^i X \vee \Sigma^i Y = \Sigma^i X \times \{ \ast \} \cup \{ \ast \} \times \Sigma^i Y \subset \Sigma^i X \times \Sigma^i Y$ consists of all cells $e \times \{ \ast \}, \{ \ast \} \times e'$ with cells $e$ of $\Sigma^i X$ and $e'$ of $\Sigma^i Y$. They have dimension $\le 2i-1$ if and only if their nontrivial factor $e$ resp. $e'$ has dimension $\le 2i-1$.
This proves the statement. Note that it can be generalized to the following:
If $X, Y$ are CW-complexes with one $0$-cell (being the base point) and no cells of dimension $1,\dots,i-1$, then $X \vee Y$ and $X \times Y$ have the same $(2i-1)$-skeleton.
Your proof is essentially correct. However, you have to use three facts:
(1) There exists a long exact sequence for the reduced homology groups, and this has to be used in your proof.
(2) $\tilde{H}_n = H_n$ for $n > 0$.
(3) $\tilde{H}_0(\ast) = 0$.
Best Answer
For the unreduced suspension for the unbased space $X$, we have \begin{align*} \Sigma X&=(X\times I)/(X\times \{0\}\cup X\times \{1\})\\ &= (X\times (0,1))\amalg \{*_0,*_1\} \end{align*} The trouble here is that we have two distinguished points $*_0$ and $*_1$ and we need to reduce it to have just one basepoint in a canonical way.
If $X$ is contractible, I guess this is what @Grumpy Parsnip meant by saying "if $X$ is nice enough" in his comment, we have $X\times (0,1)\simeq (X\times (0,1))\amalg (0,1)$. Hence, \begin{align*} \Sigma X\vee S^1&= (X\times (0,1))\amalg (0,1)\amalg \{1\sim *_0 \sim *_1\}\\&\simeq(X\times (0,1))\amalg \{*\}\\&= (X_+\times S^1)/\{*\}\times S^1 \cup X_+\times \{1\}\\ &= X_+\wedge S^1 = \Sigma X_+ \quad \text{ reduced suspension} \end{align*} The author seems to state this just to provide an example that $\Sigma (X_+)\nsim (\Sigma X)_+$ unlike a cone $CX$ where $C(X_+)\simeq(CX)_+$ so having this just for $X$ contractible seems to be enough to serve its goal.
I do realize that I'm abusing notations by ignoring topological structures by only considering a set structure level but I hope this would not be a big trouble. Let me know if anyone finds any flaw in my argument. Thanks!