I would like to know whether the following is false or not, since I can't really get a good grip on it.
Suppose $\mathcal{C}$ is any category and let $P$ be a product of two given objects $X$ and $Y$. Suppose $P'$ is an object such that $P' \cong P$. Is $P'$ also a product of $X$ and $Y$ then?
My first thinking is no, but I don't know how to prove it exactly.
Any hint is much appreciated.
I see you've already accepted an answer, but contrary to what the answer you accepted suggests, the actual answer to your question is in fact yes. The same is true of arbitrary limits (and colimits), but let's focus on products here.
Let $X \xleftarrow{\pi_1} P \xrightarrow{\pi_2} Y$ be a product diagram in $\mathcal{C}$ and suppose that $P \cong P'$. Let $u : P \to P'$ be the given isomorphism. It is easy to check that $$X \xleftarrow{\pi_1 \circ u^{-1}} P' \xrightarrow{\pi_2 \circ u^{-1}} Y$$ is also a product diagram in $\mathcal{C}$. Indeed, given morphisms $f : A \to X$ and $g : A \to Y$, the unique morphism $A \to P'$ factoring through the new product projections is precisely $u \circ \langle f,g \rangle$, where $\langle f,g \rangle$ is the unique morphism $A \to P$ factoring through the old product projections.
More generally, given any diagram $d : \mathcal{J} \to \mathcal{C}$ and any limit cone $$(L, (\lambda_j : L \to d(j))_{j \in \mathrm{ob}(\mathcal{J})})$$ for $d$ in $\mathcal{C}$, if $L \cong L'$ via an isomorphism $u : L \to L'$ in $\mathcal{C}$, then $$(L', (\lambda_j \circ u^{-1} : L' \to d(j))_{j \in \mathrm{ob}(\mathcal{J})})$$ is another limit cone for $d$ in $\mathcal{C}$, thus exhibiting $L'$ as a limit of $d$.