The spaces $\Sigma\mathbb{R}P^3$ and $\mathbb{R}P^4/\mathbb{R}P^1$ have isomorphic cohomology groups I need to prove the statement of the title, but I get different results. For $\Sigma \mathbb{R}P^3$, we know $H^{n}(\Sigma \mathbb{R}P^3) \cong H^{n - 1}(\mathbb{R}P^3)$ by Mayer-Vietoris and $\mathbb{Z}$ for $n = 0$ because there is one connected component. So the cohomology groups are $\mathbb{Z}, \mathbb{Z}, 0, \mathbb{Z}/2, \mathbb{Z}$.
For $\mathbb{R}P^4/\mathbb{R}P^1$ we get the cellular chain complex
$$\mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \to 0 \to \mathbb{Z}$$
because we collapse the $1$-skeleton and the attachment of the $3$-cell and $4$-cell don't change. So the cellular co-chain complex becomes
$$\mathbb{Z} \to 0 \to \mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{2} \mathbb{Z} $$
which means that the cohomology groups are $\mathbb{Z}, 0, \mathbb{Z}, 0, \mathbb{Z}/2$. 
These do not coincide, so where am I going wrong?
 A: With $\mathbb{Z}$ coefficients the statement you want isn't true (compare the spaces' $H^4$ for example), but it is true with $\mathbb{Z}/2$ coefficients. Your result for the quotient space looks correct, but for the suspension it's not quite accurate.
You have to be careful with the suspension isomorphism because it's actually a statement about reduced cohomology, i.e. for all $X$ and for all $n > 0$ we have $\tilde{H}^n(\Sigma X) \cong \tilde{H}^{n-1}(X)$. The statement "$H^n(\Sigma X) \cong H^{n-1}(X)$ for $n > 0$" is actually false for every space (see below for more details), think about what happens when $X=S^k$ and $\Sigma X \cong S^{k+1}$.
For $\mathbb{RP}^4/\mathbb{RP}^1$, if you wanted you could compute the cohomology without explicitly describing the whole co-chain complex, using the fact that $(\mathbb{RP}^4, \mathbb{RP}^1)$ is a "good pair" so the reduced cohomology is $\tilde{H}^n(\mathbb{RP}^4/ \mathbb{RP}^1)\cong H^n(\mathbb{RP}^4, \mathbb{RP}^1)$.
If two spaces have the same reduced cohomology then they have the same absolute cohomology, so in principle you could solve this problem by just computing reduced cohomology groups. But, again, it seems that the statement is false with $\mathbb{Z}$ coefficients.

Edit: To see what goes wrong with the statement $H^{n}(\Sigma X) \cong H^{n-1}(X)$, consider the case that $n=1$. If you decompose $\Sigma X$ into a "lower cone" $U$ and an "upper cone" $V$ (as in your comment to this answer) then we get a Mayer-Vietoris sequence, which near degree $0$ takes the form
$$H^0 (\Sigma X) \to H^0 (U)\oplus H^0 (V) \to H^0 (U\cap V) \to H^1 (\Sigma X) \to 0 $$
Recall that $H^0(Y)$ is the group of locally constant functions $Y \to \mathbb{Z}$, and in particular it is a free abelian group whose rank is the number of connected components. Using the facts that $\Sigma X$ is connected, $U$ and $V$ are contractible, and $U\cap V \simeq X$, the sequence becomes $$\mathbb{Z} \to \mathbb{Z}\oplus \mathbb{Z} \to \mathbb{Z}^k \to H^1(\Sigma X) \to 0 $$
where $k$ is the number of components of $X$. The first two homomorphisms are induced by inclusion, specifically $H^0(\Sigma X) \to H^0 (U)\oplus H^0 (V)$ sends $[f]$ to $([f|_U], [f|_V])$; the map $H^0 (U)\oplus H^0 (V) \to H^0 (U\cap V)$ sends $([f], [g])$ to $[f|_{U\cap V}] - [g|_{U\cap V}]$, and if $U\cap V$ has $k$ components then this homomorphism is surjective onto the diagonal $\Delta \subset \mathbb{Z}^k$. Therefore this sequence breaks up into two pieces:
$$\mathbb{Z} \to \mathbb{Z}\oplus \mathbb{Z} \to \Delta\text{  and  }  \mathbb{Z}^k /\Delta \cong \mathbb{Z}^{k-1} \stackrel{\cong}{\to} H^1(\Sigma X).$$
I.e. for any space $X$ the group $H^1(\Sigma X)$ is free abelian and its rank is the number of components of $X$ minus one, in other words it is isomorphic to $\tilde{H}^0(X)$. Also note that this first sequence is disregarded when we take reduced cohomology.
