You don't say how much structure you have proven for abelian groups, so I will not assume much. If you do know some structure theorems, please let us know. (E.g., there is a theorem that if $G$ is abelian, and $a$ is an element of $G$ of maximal order, then there is a subgroup $H$ of $G$ such that $G = H\oplus \langle a\rangle$; do you know that?) Anyway...
Let $G$ and $H$ be finite abelian $p$-groups that have the same number of elements of each order. We want to prove that $G$ and $H$ are isomorphic.
Let $p^n$ be the largest order of an element of $G$ (and of $H$). If $n=1$, then $G$ and $H$ are elementary abelian $p$-groups; so they are vector spaces over $\mathbf{F}_p$, and since they have the same number of elements, they are isomorphic (same dimension).
Assume the result holds for abelian groups whose largest orders are $p^k$, and let $G$ and $H$ be groups satisfying our hypothesis and in which the largest elements have oder $p^{k+1}$.
Show that $pG$ and $pH$ have the same number of elements of each order, and that the elements of largest orders have order $p^k$. Apply induction to conclude $pG\cong pH$. Now see if you can leverage that to get $G\cong H$. If you need more help with those steps, please ask through comments.
This line seems especially mistaken: "I think, by definition of a normal subgroup, they are abelian and so this tells us that G is abelian." Certainly normal subgroups need not be abelian: for an example you can take the alternating subgroup of the symmetric group for any $n>5$.
The Sylow theorems tell you that $n_7\in \{1,5,25\}$ and that it is 1 mod 7, and so the only possibility is that it is 1.
The Sylow theorems tell you that $n_5\in \{1,7\}$ and that it is 1 mod 5, and so the only possibility is that it is 1.
Thus for both 5 and 7 you have unique (=normal for Sylow subgroups) subgroups. Let's call them $F$ and $S$ respectively. Clearly $FS$ is a subgroup of $G$ of size 175 by the reasoning you gave. (The reason that $F\cap S$ is trivial is that the intersection is a subgroup of both $F$ and $S$, so it must have order dividing both the order of $F$ and of $S$, but the greatest common divisor is 1.)
$S$ is obviously abelian, as it is cyclic (of prime order!). The question is whether or not a group of size 25 must be abelian. There are a lot of ways to see that, but the one that comes to my mind is to say that it definitely has a nontrivial center. If its center $C$ were of order $5$, then $F/C$ would be cyclic of order 5. However, by a lemma (If $G/Z$ is cyclic for a central subgroup $Z$, then $G$ is abelian) $F$ would have to be abelian.
So $G$ is a product of two abelian subgroups, and so is abelian itself.
And also, your conclusion about the two types of abelian groups of order 175 is correct. Initially you wrote that there were "two isomorphic types," but (I edited that to correct it and ) I hope that was just a slip and that you really did mean "two non-isomorphic types".
Best Answer
We can certainly prove a baby version of the structure theorem for finite abelian groups:
Theorem: A finite abelian group $G$ is the direct product of its Sylow $p$-subgroups.
Proof: Let $\#G=n=p_1^{n_1}\dots p_r^{n_r}$. By Euclid/Bezout/..., there exists integers $m_1,\dots,m_r$ such that $n/p_i^{n_i}$ divides $m_i$ and $m_i\equiv 1\pmod{p_i^{n_i}}$. Then $m_1+m_2+\dots+m_r\equiv 1\pmod{p_i^{n_i}}$ for all $i$ and hence $\equiv 1\pmod{n}$. By Cauchy, $g^n=1$ for all $g\in G$ so $g^{m_i}$ has $p$-power order, i.e., lies in the Sylow-$p_i$ of $G$, and $$ g=g^{m_1+m_2+\dots+m_r}=g^{m_1}g^{m_2}\dots g^{m_r} $$ gives the desired result. QED.