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For the simplest case that $r,s,p,q$ are single elements, it is shown in these notes (page 7) how the first identity (known as the Desnanot-Jacobi identity) follows from the second identity.

Apply the second identity to the matrix

we thus arrive at the first identity, illustrated graphically as

_source

Update: Since I could not find the first identity in the literature, for the more general case when $r,s,p,q$ each contain more than a single element, I tried to check it for an example. I took $n=6$, $r=1,2$, $s=5,6$, $p=1,2$, $q=5,6$. For the $6\times 6$ matrix $A$ I took _{$$A=\left( \begin{array}{cccccc} 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 2 & 0 & 2 & 2 & 0 & 0 \\ 2 & 1 & 0 & 0 & 2 & 1 \\ 2 & 0 & 2 & 1 & 0 & 2 \\ 1 & 0 & 0 & 0 & 2 & 2 \\ \end{array} \right)$$} The left-hand-side of the first identity is 0, but for the right-hand-side I find a nonzero answer:

Incidentally, I did find a determinantal identity of a somewhat similar form in Tao's blog (last equation, Karlin's identity). But it is not quite of the form of the first identity in the OP.

For the special case that $r,s,p,q$ are single elements, it is shown in these notes (page 7) how the first identity (known as the Desnanot-Jacobi identity) follows from the second identity.

Apply the second identity to the matrix

we thus arrive at the first identity, illustrated graphically as

_source

Update: Since I could not find the first identity in the literature, for the more general case when $r,s,p,q$ each contain more than a single element, I tried to check it for an example. I took $n=6$, $r=1,2$, $s=5,6$, $p=1,2$, $q=5,6$. For the $6\times 6$ matrix $A$ I took _{$$A=\left( \begin{array}{cccccc} 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 2 & 0 & 2 & 2 & 0 & 0 \\ 2 & 1 & 0 & 0 & 2 & 1 \\ 2 & 0 & 2 & 1 & 0 & 2 \\ 1 & 0 & 0 & 0 & 2 & 2 \\ \end{array} \right)$$} The left-hand-side of the first identity is 0,
_{$\det A \det \left( \begin{array}{cc} 2 & 2 \\ 0 & 0 \\ \end{array} \right) = 24\cdot 0 = 0$,}
but for the right-hand-side I find a nonzero answer:

Incidentally, I did find a determinantal identity of a somewhat similar form in Tao's blog (last equation, Karlin's identity). But it is not quite of the form of the first identity in the OP.

So unless I have made a mistake, my conclusion is that the first identity in the OP only holds when $r,s,p,q$ are single elements, but not more generally.

For the simplest case that $r,s,p,q$ are single elements, it is shown in these notes (page 7) how the first identity (known as the Desnanot-Jacobi identity) follows from the second identity.

Apply the second identity to the matrix we thus arrive at the first identity, illustrated graphically as _source

Update: Since I could not find the first identity in the literature, for the more general case when $r,s,p,q$ each contain more than a single element, I tried to check it for an example. I took $n=6$, $r=1,2$, $s=5,6$, $p=1,2$, $q=5,6$. For the $6\times 6$ matrix $A$ I took $$A=\left( \begin{array}{cccccc} 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 2 & 0 & 2 & 2 & 0 & 0 \\ 2 & 1 & 0 & 0 & 2 & 1 \\ 2 & 0 & 2 & 1 & 0 & 2 \\ 1 & 0 & 0 & 0 & 2 & 2 \\ \end{array} \right)$$ The left-hand-side of the first identity is 0, but for the right-hand-side I find a nonzero answer ($-20$):

Incidentally, I did find a determinantal identity of a somewhat similar form in Tao's blog (last equation, Karlin's identity). But it is not quite of the form of the first identity in the OP.

For the simplest case that $r,s,p,q$ are single elements, it is shown in these notes (page 7) how the first identity (known as the Desnanot-Jacobi identity) follows from the second identity.

Apply the second identity to the matrix 

we thus arrive at the first identity, illustrated graphically as

_source

Update: Since I could not find the first identity in the literature, for the more general case when $r,s,p,q$ each contain more than a single element, I tried to check it for an example. I took $n=6$, $r=1,2$, $s=5,6$, $p=1,2$, $q=5,6$. For the $6\times 6$ matrix $A$ I took _{$$A=\left( \begin{array}{cccccc} 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 2 & 0 & 2 & 2 & 0 & 0 \\ 2 & 1 & 0 & 0 & 2 & 1 \\ 2 & 0 & 2 & 1 & 0 & 2 \\ 1 & 0 & 0 & 0 & 2 & 2 \\ \end{array} \right)$$} The left-hand-side of the first identity is 0, but for the right-hand-side I find a nonzero answer:

Incidentally, I did find a determinantal identity of a somewhat similar form in Tao's blog (last equation, Karlin's identity). But it is not quite of the form of the first identity in the OP.

For the simplest case that $r,s,p,q$ are single elements, it is shown in these notes (page 7) how the first identity (known as the Desnanot-Jacobi identity) follows from the second identity.

Apply the second identity to the matrix we thus arrive at the first identity, illustrated graphically as _source

Update: Since I could not find the first identity in the literature, for the more general case when $r,s,p,q$ each contain more than a single element, I tried to check it for an example. I took $n=6$, $r=1,2$, $s=5,6$, $p=1,2$, $q=5,6$. For the $6\times 6$ matrix $A$ I took $$A=\left( \begin{array}{cccccc} 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 2 & 0 & 2 & 2 & 0 & 0 \\ 2 & 1 & 0 & 0 & 2 & 1 \\ 2 & 0 & 2 & 1 & 0 & 2 \\ 1 & 0 & 0 & 0 & 2 & 2 \\ \end{array} \right)$$ The left-hand-side of the first identity is 0, but for the right-hand-side I find a nonzero answer ($-20$). Have I made a mistake?

Incidentally, I did find a determinantal identity of a somewhat similar form in Tao's blog (last equation, Karlin's identity). But it is not quite of the form of the first identity in the MO.

For the simplest case that $r,s,p,q$ are single elements, it is shown in these notes (page 7) how the first identity (known as the Desnanot-Jacobi identity) follows from the second identity.

Apply the second identity to the matrix we thus arrive at the first identity, illustrated graphically as _source

Update: Since I could not find the first identity in the literature, for the more general case when $r,s,p,q$ each contain more than a single element, I tried to check it for an example. I took $n=6$, $r=1,2$, $s=5,6$, $p=1,2$, $q=5,6$. For the $6\times 6$ matrix $A$ I took $$A=\left( \begin{array}{cccccc} 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 2 & 0 & 2 & 2 & 0 & 0 \\ 2 & 1 & 0 & 0 & 2 & 1 \\ 2 & 0 & 2 & 1 & 0 & 2 \\ 1 & 0 & 0 & 0 & 2 & 2 \\ \end{array} \right)$$ The left-hand-side of the first identity is 0, but for the right-hand-side I find a nonzero answer ($-20$):

Incidentally, I did find a determinantal identity of a somewhat similar form in Tao's blog (last equation, Karlin's identity). But it is not quite of the form of the first identity in the OP.