Let $G$ be a finite group that can be generated by $2$ elements, and let $H \leq G$ be a (not necessarily normal) subgroup for which there exists some $g \in G$ such that $H \langle g\rangle = G$. Must there be $h \in H$, and $a \in G$ such that $\langle h,a \rangle = G$ ?
NOTE: $H\langle g \rangle = \{hg^n | h \in H, n \in \mathbb{Z}\}$ which is NOT the same as $\langle H,g \rangle$ which denotes the subgroup generated by $H$ and $g$.
I think I can modify Peter Mueller's example to make it a counterexample to the question that was asked. We still let $H$ be elementary abelian of order $4$, but now we take $V = C_{105} \cong C_3 \times C_5 \times C_7$. Again we let the three involutions of $H$ act as diagonal matrices on the three prime order direct factors of $V$ as $(-1,-1,1)$, $(-1,1,-1)$, and $(1,-1,-1)$, and set $G = V \rtimes H$.
Then again $G$ is $2$-generated and $\langle h,g \rangle \ne G$ for all $h \in H$, $g \in G$, but now $V = \langle g \rangle$ is cyclic and $G=HV$.
To show that $G$ is $2$-generated, let $a,b$ be the generators of $H$ that induce $(-1,-1,1)$ and $(-1,1,-1)$, and let $x,y,z$ be the generators of the cyclic subgroups of $G$ of orders $3,5,7$. We claim that $G = \langle az, bxy \rangle$. Let $F = \langle az,bxy \rangle$. Then $z$ is a power of $az$ and $y$ is a power of $bxy$, so $y,z \in F$. Also, since $[a,x]$ is a power of $x$, $[az,bxy] = x^iy^jz^k$ with $i \ne 0 \mod 3$, so $x \in F$, and now it is clear that $F=G$.
We claim also that $G$ is not generated by $h,g$ with $h \in H$, $g \in G$. Suppose that it is, and assume WLOG that $h=a$ and $g=bv$ with $v = x^iy^jz^k \in V$. Note that $bz^k$ is conjugate to $a$ and commutes with $a$, but then $\langle a,bv \rangle$ is contained in the proper subgroup $\langle a,bz^k,x,y \rangle$, contradiction.
In case anyone wants to check the calculation by computer, the group $G$ is $\mathtt{SmallGroup}(420,30)$ with $H \in \mathrm{Syl}_2(G)$.
There was a misunderstanding: the example below is NOT a counterexample.
Another example is $G=A_5^4$ with $H = \langle (t,1,1,1),(1,t,1,1),(1,1,t,1),(1,1,1,t) \rangle$, where $t=(1,2)(3,4)$.
I won't give a detailed explanation, but the reason is that, of the $19$ ordered pairs of generators of $A_5$ that are pairwise distinct up to automorphisms of $A_5$, only three have their first component of order $2$, so $G$ cannot be generated by an element of $H$ and one other element. On the other hand it is easy to generate $A_5^4$ with $H$ and one other element.
