The OP's description of the strength is correct, but there's a simpler one: just think of it as corresponding to the composite

$$A \stackrel{coev_A}{\to} \hom(B, A \times B) \to \hom(T(B), T(A \times B))$$

under the adjunction $(- \times T(B)) \dashv \hom(T(B), -)$. Here $coev$ denotes the unit of the adjunction $(- \times B) \dashv \hom(B, -)$, and the second arrow comes from $T$'s being a ($Set$-enriched) functor.

Since we are dealing with functors on $Set$, we may as well make this very concrete and describe this in terms of sets and elements (we remember sets and elements, don't we? (-: ). One can work out as an exercise that the strength on $T$ according to the description above is

$$\langle a \in A, y \in T(B)\rangle \mapsto T(b \in B \mapsto \langle a, b\rangle)(y)$$

where on the right we are applying the functor $T$ to the indicated function of type $B \to A \times B$ induced by an element $a \in A$.

(The categorical definition above generalizes so that given a closed monoidal category $V$, the structure of a $V$-monoidal strength on a functor $T: V \to V$ is equivalent to the structure of a $V$-enrichment on $T$. This doesn't require dealing with any colimits involved in coends; it just uses $V$ being closed monoidal.)

To connect this description with the OP's description, it help to be describe concretely what the co-Yoneda isomorphism is doing, again in terms of sets and elements. The co-Yoneda isomorphism

$$\int^C \hom(C, A) \times T(C) \stackrel{\sim}{\to} T(A)$$

takes a typical element given by an equivalence class $[f: C \to A, x \in T(C)]$ to $T(f)(x) \in T(A)$.

There are actually two co-Yoneda isomorphisms implicit in the OP's description:

$$\int^{C, D} \hom(1, C) \times T(D) \times \hom(C \times D, A \times B) \cong \int^D T(D) \times \hom(1 \times D, A \times B)$$

$$\int^D T(D) \times \hom(D, A \times B) \cong T(A \times B)$$

Putting these together, the composite co-Yoneda isomorphism can be described as

$$[c \in C, y \in T(D), g: C \times D \to A \times B] \qquad \mapsto \qquad T(g(c, -))(y)$$

Now setting $A = C, B = D$, and $g = id_{A \times B}$, we easily calculate the OP's composite as the function

$$\langle a \in A, y \in T(B) \rangle \mapsto T(id_{A \times B}(a, -))(y)$$

where $id_{A \times B}(a, -)$ is just another way of writing $b \mapsto \langle a, b\rangle$. Hence the two descriptions are equivalent.