I tried to prove a surjective function $f:\Bbb R\to\Bbb R$, symmetric with respect to the $II\ \&\ IV$ quadrant bisector, has to be bijective.
First, we take a restriction $\overset{\sim}{f}_{|\ [0,\infty)}$ of an arbitrary injective function $\overset{\sim}{f}$ (whose graph passes through the origin).
Then we take an inverse function $\overset{\sim}{f}^{-1}_{\overset{\sim}{f}([0,+\infty))}$ and appended a negative part to its graph so as to get an odd function that is symmetric with respect to the origin.
The range of the inverse function expanded.
An arbitrary function and its inverse are symmetric with respect to the $I\ \&\ III$ quadrant bisector.
$$T(x,f(x))\mapsto T'(f(x),x)$$
But, since the range of the inverse function expanded:
$$T'(\overset{\sim}{f}(x),x)\ \land\ -[T'(\overset{\sim}{f}(x),x)]\ \in\Gamma_{\overset{\sim}{f}^{-1}}$$ $$-T'(\overset{\sim}{f}(x),x)=(-\overset{\sim}{f}(x),-x)$$
The negative part of the extended inverse function is now appended to the first function $f.$
$$P(x,f(x))\in \Gamma_f\ \land\ P'(-f(x),-x)\in \Gamma_f\implies$$ $$(\forall x\in \Gamma_f)(\exists (-x)\in \mathcal R_f)$$
Therefore, a surjective function symmetric with respect to the (straight) line $y=-x$, i.e.$\ II\ \&\ IV$ quadrant bisector has to be bijective.
Is this a legitimate proof?
I know you're looking for proof verification, so forgive me for giving another approach: Given $(a,b)$ in the plane, what is the symmetric point about the line $y=-x?$ It is the point $(-b,-a).$
Now suppose $a\ne b$ and $f(a)=f(b)= c.$ Then $(a,c)$ and $(b,c)$ are on the graph of $f.$ Since that graph is symmetric with respect to the line $y=-x,$ the points $(-c,-a)$ and $(-c,-b)$ are also on the graph. This implies $f(-c) = -a$ and $f(-c) = -b.$ But that is impossible for the graph of a function. It follows that $a\ne b$ implies $f(a)\ne f(b).$ Thus $f$ is injective. Since $f$ is given to be surjective, $f$ is a bijection.