We know the cohomology ring with coefficients in $\mathbb{Z}/2$ of projective real space $\mathbb{PR}^n$ is
$$H^*(\mathbb{PR}^n, \mathbb{Z}/2) = \mathbb{Z}/2[X]/(X^{n+1})$$
with graduated generator $X \in H^1(\mathbb{PR}^n, \mathbb{Z}/2)$. Futhermore, by Künneth for cohomology we also have isomomorpism of graduated rings
$$H^*(\mathbb{PR}^n \times \mathbb{PR}^m, \mathbb{Z}/2) \cong H^*(\mathbb{PR}^n, \mathbb{Z}/2) \otimes H^*(\mathbb{PR}^m, \mathbb{Z}/2) \cong \mathbb{Z}/2[a]/(a^{n+1}) \otimes \mathbb{Z}/2[b]/(b^{m+1}) \cong \mathbb{Z}/2[a, b]/(a^{n+1}, b^{n+1})$$
the last one comes with identifications $a \otimes 1 \mapsto a, 1 \otimes b \mapsto b$.
From now we consider $\mathbb{PR}^n \cong (\mathbb{PR}^n, pt)$ as pointed space. Therfore we can define following canonical morphisms:
$i_n:\mathbb{PR}^n \times \{pt\} \hookrightarrow \mathbb{PR}^n \times \mathbb{PR}^m $
$pr_n: \mathbb{PR}^n \times \mathbb{PR}^m \to \mathbb{PR}^n \times \{pt\}$.
Analogously $i_m, pr_m$ for $\{pt\} \times \mathbb{PR}^m$. Let us now consider the induced morphisms $i_n^*$ and $pr_n^*$ on cohomology. Cince $pr_m \circ i_n, pr_n \circ i_m$ are constant maps and $pr_n \circ i_n = id_{\mathbb{PR}^n}$, $pr_m \circ i_m = id_{\mathbb{PR}^m}$ und $H^*(-)$ is functorial the compostions above become zero maps resp identity on corresponding cohomology rings.
My question is the following:
I don't know how to use the given identity $ i_k ^* \circ pr_l ^* = \delta_{k,l} Id $ for $k,l \in \{m,n\}$
to derive explicitely that
$i_n ^*:H^*(\mathbb{PR}^n \times \mathbb{PR}^m, \mathbb{Z}/2) \to H^*(\mathbb{PR}^n , \mathbb{Z}/2)$ is given by
$a \otimes 1 \to a, 1 \otimes b \to 0$
and
$pr_n ^*:H^*(\mathbb{PR}^n , \mathbb{Z}/2) \to H^*(\mathbb{PR}^n \times \mathbb{PR}^m, \mathbb{Z}/2)$ by
$a \to a \otimes 1$.
My attempts:
$i_n^*, pr_m^*$ respect grades and therefore $pr_n^*$ maps $$a \to d_n a \otimes 1 + e_n 1 \otimes b$$
with unknown $d_n, e_n \in \mathbb{Z}/2$. Compose with $i_n^*$ gives $a = i^*_n \circ pr_n^*(a) = i^*_n(d_n a \otimes 1 + e_n 1 \otimes b) = d_n i^*_n(a \otimes 1) + e_n i^*_n(1 \otimes b)$ and $0 = i^*_m \circ pr_n^*(a) = i^*_n(d_n a \otimes 1 + e_n 1 \otimes b) = d_n i^*_n(a \otimes 1) +e_n i^*_n(1 \otimes b)$.
I get analogous system applying $pr_m ^*$ with another coefficients $d_m, e_m$. From here - without extra information - I'm stuck. Could anybody help?
I think your misunderstanding stems from the fact that you are using the Künneth Theorem as an identification. You are writing $a\otimes b$ as an element of $H^*(\mathbb{R}P^m\times\mathbb{R}P^n;\mathbb{Z}_2)$, which it is not. Actually $a\otimes b\in H^*(\mathbb{R}P^m;\mathbb{Z}_2)\otimes_{\mathbb{Z}_2}H^*(\mathbb{R}P^n;\mathbb{Z}_2)$, so it does not make sense to apply the homomorphism $pr_1^*:H^*(\mathbb{R}P^m\otimes\mathbb{R}P^n;\mathbb{Z}_2)\rightarrow H^*(\mathbb{R}P^n;\mathbb{Z}_2)$ to this element, since it does not have the correct domain.
To see what is going on, you need to understand that the Künneth Isomorphism is induced by the cohomology cross product. For example, Hatcher, on page 214, defines the cross product (or external cup product) for cohomology over a ground ring $R$
$$K:H^*X\otimes_R H^*Y\rightarrow H^*(X\times Y)$$
by
$$a\otimes y\mapsto x\times y=pr_1^*x\cup pr_2^*y\in H^{m+n}(X\times Y),\qquad x\in H^mX,y\in H^nY,$$
where $X,Y$ are spaces and $pr_1:X\times Y\rightarrow Y$, $pr_2:X\times Y\rightarrow Y$ are the projections. A version of the Künneth Theorem appears on page 219 as Theorem 3.18 and states that the cross product $K$ is an isomorphism of rings if $H^*Y$ is a finitely generated free graded $R$-module. Certainly these conditions hold for $X=\mathbb{R}P^m$, $Y=\mathbb{R}P^m$ and $R=\mathbb{Z}_2$.
Thus you really apply $pr_1^*$ to the element $K(a\otimes b)=a\times b=pr_1^*a\cup pr_2^*b$. So as it turns out the second of your queries is tautologically true if you are willing to assume the Künneth formula as given, and that $i_n^*(a\times 1)=1$ then follows directly from your computations, since $(a\times 1)=pr_1^*a\cup pr_2^*1=pr_1^*a\cup 1=pr_1^*a$. Note the fact that $pr_1^*1=1$ follows from the fact the cross product is an isomorphism of rings.