The second question has a well known, affirmative answer from invariant theory.

Note that $O(V)$ acts on every tensor power $V^{\otimes k}$ and by conjugation also on $Hom(V^{\otimes a},V^{\otimes b})$ and the homomorphisms commuting with $O(V)$ are exactly the fixed points of this conjugation action. We have a natural isomorphism
$$Hom(V^{\otimes a},V^{\otimes b}) \cong (V^{\otimes a})^\ast \otimes V^{\otimes b} \cong (V^\ast)^{\otimes a}\otimes V^{\otimes b}$$
The scalar product gives us an isomorphism $V\cong V^\ast$ which translates this into
$$Hom(V^{\otimes a},V^{\otimes b}) \cong V^{\otimes a}\otimes V^{\otimes b} = V^{\otimes(a+b)}$$
Explicitly this last one is given by
$$x_1\otimes\cdots\otimes x_a \otimes y_1\otimes \cdots \otimes y_b \mapsto \left(v_1\otimes\cdots\otimes v_a \mapsto \prod_{i=1}^a \langle x_i,v_i\rangle y_1\otimes\cdots\otimes y_b\right)$$
All of these are natural w.r.t. isometries so that our problem translates into finding fixed points on $V^{\otimes k}$ for every $k\in\mathbb{N}$. Since $v\mapsto -v$ is an isometry, there are no non-zero fixed points for uneven $k$. So we are left with the case where $a+b$ is even.

For the special case $a=b$, we want the centraliser of $\rho(O(V))$ inside $End(V^{\otimes a})$ where $\rho$ is the representation $O(V)\to End(V^{\otimes a}), \phi\mapsto\phi^{\otimes a}$. This centraliser turns out to be well-known: It is $\psi(B_a(d))$ where $d:=\dim(V)$, $B_a(d)$ is the so-called Brauer algebra and $\psi: B_a(d) \to End(V^{\otimes a})$ is its natural representation. This is the analogue of Schur-Weyl-duality for orthogonal groups.

The Brauer algebra has a basis indexed by perfect pairings $D$ of the numbers $1,2,\ldots,2a$ each of which describes a combination of permutations (when a number in $\{1,\ldots,a\}$ is paired with a number in $\{a+1,\ldots,2a\}$) and the map $v_1\otimes v_2 \mapsto \langle v_1,v_2\rangle \Omega$, where $\Omega:=\sum_{i=1}^{\dim(V)} e_i\otimes e_i$ is the Casimir element.

If we follow the isomorphisms above, we find that these basis elements give us the following spanning set of fixed points in $(V^{\otimes 2a})^{O(V)}$:
$$\Omega_D := \sum_{\substack{1\leq i_1,\ldots,i_{2a}\leq \dim(V) \\ \lbrace x,y\rbrace\in D \implies i_x=i_y}} e_{i_1} \otimes e_{i_2} \otimes \cdots \otimes e_{i_{2a}}$$
where $D$ runs over all perfect pairings of the numbers $1,2,\ldots,2a$.

Now the fixed points do not care that arrived at the number $2a$ by considering $a=b$. Therefore this is also the answer for the general case where $a+b$ is any even number. If we follow all the isomorphisms backward, we find that these fixed points $\Omega_D\in V^{\otimes(a+b)}$ represent the maps $\tau^D: V^{\otimes a}\to V^{\otimes b}$ that come from permuting some tensor factors (=numbers in $\{1,\ldots,a\}$ get paired with numbers in $\{a+1,\dots,a+b\}$), contracting some (=two numbers in $\{1,\ldots,a\}$ get paired with each other), and introducing Casimir elements in some factors of the result (=two numbers in $\{a+1,\ldots,a+b\}$ get paired with each other) which is what we wanted to prove.

To prove the original conjecture, we have to find all the linear combinations
$$\tau = \sum_D \alpha_D \tau^D$$
which are natural not only w.r.t to bijective isometries but w.r.t. all isometries. Our theorem is proven if we can prove that
$$D\text{ contains a pairing }\{\alpha,\beta\} \subseteq \{a+1,\ldots,a+b\} \implies \alpha_D=0$$
Because then all introductions of Casimir elements are eliminated and only traces and reordering of tensor factors remain. In particular it would follow that $a\geq b$ must hold if $\tau\neq 0$. That together with the earlier parity argument would prove our theorem.

Step 1.:
Consider a vector space $V$ of dimension $d$ with an orthonormal basis $e_1,\ldots,e_d$ and define for any assignment of indices $j: \{1,\ldots,a\} \to \{1,\ldots,d\}$ the tensor $v_j\in V^{\otimes a}$ as
$$v_j := e_{j(1)} \otimes e_{j(2)} \otimes \cdots \otimes e_{j(a)}$$

If we have two such assignments $j: \{1,\ldots,a\} \to \{1,\ldots,d\}$ and $k: \{1,\ldots,b\} \to \{1,\ldots,d\}$, then we can compute the scalar product $\langle{\tau_V(v_j),v_k}\rangle$ as follows:
\begin{align*}
\langle{\tau_V(v_j),v_k}\rangle &= \sum_D \alpha_D \langle{\tau_V^D(v_j),v_k}\rangle \\
&= \sum_{D,i} \alpha_D \prod_{s=1}^a \langle{e_{i(s)},e_{j(s)}}\rangle \langle{e_{i(a+1)}\otimes\cdots\otimes e_{i(a+b)}, e_{k(1)}\otimes\cdots\otimes e_{k(b)}} \rangle \\
&= \sum_{D,i} \alpha_D \prod_{s=1}^a \langle{e_{i(s)},e_{j(s)}}\rangle \prod_{t=1}^b \langle{e_{i(a+t)}, e_{k(t)}}\rangle
\end{align*}

where - as above - $i$ runs over all indexing functions $i:\{1,2,\ldots,a+b\}\to\{1,\ldots,d\}$ such that
$$\{x,y\}\in D \implies i(x)=i(y).\tag{*}$$

Step 2: We prove the following lemma

If $d\geq \frac{a+b}{2}$, then for every perfect pairing $D$, there are assignments $j: \{1,\ldots,a\} \to \{1,\ldots,d\}$ and $k: \{1,\ldots,b\} \to \{1,\ldots,d\}$ such that
$$\langle{\tau_V(v_j),v_k}\rangle = \alpha_D$$
If moreover $D$ contains a pair $\{\alpha,\beta\} \subseteq \{a+1,\ldots,a+b\}$, then it is possible to find such $j$ and $k$ with $\forall 1\leq x\leq a: j(x) \leq m-1$.

Proof of the Lemma:
To see this, note that the factors in both products in the above expression for the scalar product are either one or zero so that if the whole summand is non-zero, then $\forall s: i(s) = j(s)$ and $\forall t: i(a+t) = k(t)$ must hold, i.e. there can only be one $i$: The one that agrees with $j$ on $\{1,\ldots,a\}$ and with $k$ on $\{a+1,\ldots,a+b\}$. Therefore most summands in the expression are zero anyway and can be ignored. The sum reduces to
$$\langle{\tau_V(v_j),v_k}\rangle = \sum_{D} \alpha_D \prod_{s=1}^a \langle{e_{i(s)},e_{j(s)}}\rangle \prod_{t=1}^b \langle{e_{i(a+t)}, e_{k(t)}}\rangle$$
where $i$ is the one function that agrees with $j$ in the lower and with $k$ in the upper part of $\{1,2,\ldots,a,a+1,\ldots,a+b\}$.

Additionally, there are only a limited number of pairings $D$ that are compatible with this indexing function $i$, because $i$ also has to satisfy the condition $(*)$.
Now let our given pairing $D$ be $\Big\{\{x_1,y_1\}, \{x_2,y_2\}, \ldots, \{x_m,y_m\}\Big\}$. Then if we choose $j$ and $k$ such that $i(x_1)=i(y_1) = 1$, $i(x_2)=i(y_2)=2$, ..., $i(x_m)=i(y_m)=m$, then there is only a single pairing that satisfies $(\ast)$ for this indexing function $i$, namely the pairing $D$ we started with.

All summands other than the $D$-th are therefore zero and the whole sum simplifies to
$$\langle{\tau_V(v_j),v_k}\rangle = \alpha_D \prod_{s=1}^a \langle{e_{i(s)},e_{j(s)}}\rangle \prod_{t=1}^b \langle{e_{i(a+t)}, e_{k(t)}}\rangle = \alpha_D\cdot 1$$

We have complete freedom which numbering of the pairs in $D$ we choose to define $i$. If $\{\alpha,\beta\} \subseteq \{a+1,\ldots,a+b\}$ is one of the pairs in $D$, we can choose our ordering such that $x_m=\alpha$ and $y_m=\beta$. Therefore the numbers $1,2,\ldots,a$ get mapped to something other than $m$ by $i$. That proves the lemma.

Now back to the proof of the theorem.

Step 3. Let's focus on a specific example: Choose $V$ to be large enough that $d\geq \frac{a+b}{2}$ and set $U:=span(e_1,\ldots,e_{m-1})$. The embedding $\phi: U\to V, v\mapsto v$ is an isometry so that our natural transformation $\tau$ must satisfy
$$\forall v\in U^{\otimes a}: \tau_U(v) = \tau_V(v)$$
Now in $\tau_U(v)$ all the sums only have indices running up to $\dim(U)=m-1$ whereas in $\tau_V(v)$ the same indices run up to $d$. On the right hand side however some summands containing $e_m,e_{m+1},\ldots,e_d$ may appear.

Consider the linear combination $\tau = \sum_D \alpha_D \tau^D$ we started with and let $D$ be one of the perfect pairings which contain a pair $\{\alpha,\beta\} \subseteq \{a+1,\ldots,a+b\}$. By the lemma just proven, there is an indexing function $i$ and its two parts $j$ and $k$ such that $j$ takes only values inside $\{1,\ldots,m-1\}$ and $\langle{\tau_V(v_j),v_k}\rangle = \alpha_D$.

Moreover, because $j$ takes values in $\{1,\ldots,m-1\}$, $v_j$ must be in $U^{\otimes a}$ and $v_k$ contains $e_m,e_{m+1},\ldots,e_d \in U^\perp$ in the $\alpha$-th (and $\beta$-th too) tensor factor. We conclude
$$\alpha_D = \langle{\tau_V(v_j),v_k}\rangle = \langle{\tau_U(v_j),v_k}\rangle = 0$$
and that proves the theorem.