= Solution
For an oriented real rank-$r$ vector bundle $\pi:E\to X$, let $u_E\in H^r(D(E),S(E);\mathbb Z)$ be its <Thom class>. If $s:X\to D(E)$ is the zero section, the <Euler class of a vector bundle> is
$$
e(E)=s^*u_E\in H^r(X;\mathbb Z).
$$
Let $u$ generate $H^{2k}(S^{2k};\mathbb Z)$. The <Poincaré-Hopf theorem> gives
$$
e(TS^{2k})=\chi(S^{2k})u=2u.
$$
For every integer $m$, choose a map $f_m:S^{2k}\to S^{2k}$ of degree $m$. Naturality of the <Euler class> gives
$$
e(f_m^*TS^{2k})=f_m^*e(TS^{2k})=2m,u.
$$
Thus every even class belongs to $\mathcal E_{2k}(S^{2k})$.
More generally, every class $a\in H^{2k}(M;\mathbb Z)$ on a closed $2k$-manifold is $f^*u$ for some map $f:M\to S^{2k}$, by the <realization of top-dimensional cohomology by a sphere map>. Pulling back $TS^{2k}$ gives
$$
2a=e(f^*TS^{2k}),
$$
so $2H^{2k}(M;\mathbb Z)\subseteq\mathcal E_{2k}(M)$. The inclusion can be strict: every <complex line bundle> on $S^2$ has an underlying oriented real two-plane bundle, and these bundles realize every integral Euler class. Hence
$$
2H^2(S^2;\mathbb Z)\subsetneq\mathcal E_2(S^2)=H^2(S^2;\mathbb Z).
$$
The odd-dimensional analogue fails. The <Euler class of an oriented odd-rank vector bundle is two-torsion>, whereas $H^{2k+1}(S^{2k+1};\mathbb Z)\cong\mathbb Z$ is torsion-free. Thus every oriented rank-$(2k+1)$ bundle on $S^{2k+1}$ has zero Euler class, and the nonzero subgroup $2\mathbb Z$ cannot lie in $\mathcal E_{2k+1}(S^{2k+1})$.
Finally, if $E$ and $F$ are oriented bundles of ranks $k$ and $l$, orient $E\oplus F$ by the ordered sum. The <Whitney product formula for Euler classes> states
$$
e(E\oplus F)=e(E)\smile e(F).
$$
Therefore cup product restricts to the asserted map $\mathcal E_k(X)\otimes\mathcal E_l(X)\to\mathcal E_{k+l}(X)$.
It need not be injective. For $X=S^2$ and $k=l=2$, one has $\mathcal E_2(S^2)=\mathbb Z$, so the source contains $\mathbb Z\otimes\mathbb Z\cong\mathbb Z$, but $H^4(S^2)=0$ and the map is zero. It need not be surjective either. For $X=S^2$ and $k=l=1$, every oriented real line bundle is trivial, so both degree-one Euler-class sets vanish, while $\mathcal E_2(S^2)=\mathbb Z$.
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