I now have a completely elementary proof. I thought I'd say a little bit about how I found this. I suspected that the only relevant properties of primes which are $7 \mod 8$ are that $2$ is a QR and $-1$ is not. Just using these two facts gave me tons of relationships between sums of Legnedre symbols, but I was getting lost in a pile of relationships without being able to pick out the ones I needed. I cut down the clutter in two ways: (1) I was frequently breaking my sums up into parts. I decided that I was only going to break the set $\{ 1,2, \ldots, p-1 \}$ into four pieces, no more, and pursue that line to the end. If it failed, I'd go back and try more pieces. (2) Since all the relations I was finding were linear, I didn't have to try to fit them into a logical chain. I just had to write down everything I knew, and what I wanted to conclude; it was then a matter of mechanical linear algebra whether or not my givens implied my conclusion.
So, we will partition $\{ 1,2, \ldots, p-1 \}$ into $4$ sets. An element $r$ is in
- $A$ if $r$ is odd and $r < p/2$
- $B$ if $r$ is even and $r < p/2$
- $C$ if $r$ is odd and $r>p/2$
- $D$ if $r$ is even and $r>p/2$.
For $X$ one of the sets $A$, $B$, $C$, $D$, write $S(X)$ for $\sum_{r \in X} \left( \frac{r}{p} \right)$ and write $T(X)$ for $\sum_{r \in X} r \left( \frac{r}{p} \right)$. We have the following relations (exercise!) $ \begin{array}{r@{}c@{}lr@{}c@{}l} S(D)&=&-S(A) & T(D) &=&- (p S(A) - T(A)) \\ S(C)&=&-S(B) & T(C)&=&- (p S(B) - T(B)) \\ S(A)+S(B)&=&S(B)+S(D) &2(T(A)+T(B)) &=& T(B)+T(D) \end{array}$
The left three equations imply that $(S(A), S(B), S(C), S(D)) = (0,h,-h,0)$ for some $h$. Then the right three imply that $(T(A),T(B),T(C),T(D)) = (x,-x,-x-ph,x)$ for some $x$. None of this required creative thought; I just found the kernel of a $6 \times 8$ matrix.
Your desired conclusion is $T(A)+T(B)=0$, which we see is true. The equality between Dirichlet's expressions is $S(A)+S(B) = -(1/p) (T(A)+T(B)+T(C)+T(D))$, which we also see is true.