Remark about square roots: I thought a bit about how to make the solution more algorithmic, and remembered that, if $\mathcal{S}_N$ is the connected open set consisting of those $N$-by-$N$ matrices without a real, nonpositive eigenvalue, then there is a canonical square root function $\sigma:\mathcal{S}_N\to \mathcal{M}_N$ ($=$ all $N$-by-$N$ complex matrices) with the property that $\sigma(A)^2 = A$ for all $A\in \mathcal{S}_N$ and that all of the eigenvalues of $\sigma(A)$ have positive real part. Using $\sigma$, we have an explicit formula for $(H_1,H_2)$ in terms of $(P_1,P_2)$, when $(P_1,P_2)$ satisfy the open condition that $Q^2{-}I$ lie in $\mathcal{S}_N$. For example, we have $$ H_2 = \tfrac12(P_2{-}I) + (I{+}P_1)^{-1}\sigma(Q^2{-}I)\ , $$ in this case. What's not completely obvious is that the expression on the right hand side is always an Hermitian symmetric matrix, but, because the second term $(I{+}P_1)^{-1}\sigma(Q^2{-}I)$ is an analytic expression in $(P_1,P_2)$ on the (nonempty) open set in $\mathcal{P}_N{\times}\mathcal{P}_N$ for which $Q^2{-}I$ lies in $\mathcal{S}_N$ and because this open set clearly has nontrivial intersection with the set $F\bigl(\mathcal{P}_N{\times}\mathcal{P}_N\bigr)$ where this expression is Hermitian symmetric, it follows that it must be Hermitian symmetric whenever $Q^2{-}I$ lies in $\mathcal{S}_N$. Thus, I believe this gives us an 'explicit' inversion of $F$ on a large open set in the image of $F$. (I put 'explicit' in quotes because you may not feel that $\sigma$ has been 'explicitly' defined.)

I assume that the problem is to try to determine which pairs $(P_1,P_2)$ of positive definite Hermitian symmetric $N$-by-$N$ matrices can be written in the above form for some pair $(H_1,H_2)$ of positive semi-definite Hermitian symmetric $N$-by-$N$ matrices.

Let $\mathcal{H}_N$ denote the set of Hermitian symmetric $N$-by-$N$ matrices, and let $\mathcal{P}_N\subset \mathcal{H}_N$ denote the open convex cone consisting of the positive definite ones, with $\mathcal{K}_N = \overline{\mathcal{P}_N}$ denoting its closure, i.e., the set of positive semidefinite ones. Finally, let $\mathcal{J}_N\subset\mathcal{H}_N$ denote the convex open set consisting of the elements $H\in \mathcal{H}_N$ such that $I{+}H$ lies in $\mathcal{P}_N$. Note that $\mathcal{P}_N\subset \mathcal{K}_N\subset \mathcal{J}_N$. Consider the smooth map $F:\mathcal{J}_N\times \mathcal{J}_N\to \mathcal{P}_N\times \mathcal{P}_N$ defined by $$ F(H_1,H_2) = \bigl(H_1 + (I{+}H_2)^{-1}, H_2 + (I{+}H_1)^{-1}\bigr). $$ A straightforward computation shows that if $(H_1,H_2)\in\mathcal{K}_N\times\mathcal{K}_N$ satisfies $H_1{+}H_2\in \mathcal{P}_N$, then $F'(H_1,H_2):\mathcal{H}_N{\oplus}\mathcal{H}_N\to \mathcal{H}_N{\oplus}\mathcal{H}_N$ is an isomorphism, so, in particular, $F$ is a local diffeomorphism on a neighborhood of $(H_1,H_2)$. In particular, the mapping $$ F:\mathcal{P}_N\times \mathcal{P}_N\to \mathcal{P}_N\times \mathcal{P}_N $$ is a local diffeomorphism, so $F\bigl(\mathcal{P}_N\times\mathcal{P}_N\bigr)$ is an open subset of $\mathcal{P}_N\times \mathcal{P}_N$. Thus, there cannot be any algebraic relations implied between $P_1$ and $P_2$ in order that there exist an $(H_1,H_2)\in\mathcal{K}_N\times\mathcal{K}_N$ so that $F(H_1,H_2) = (P_1,P_2)$. The most one can hope for is inequalities relating $P_1$ and $P_2$ that might characterize the image $F\bigl(\mathcal{K}_N\times\mathcal{K}_N\bigr)$ in $\mathcal{P}_N\times\mathcal{P}_N$.

One inequality is very easy, and turns out to be quite useful: Note that $$ \begin{align} P_1 &= H_1 + (I{+}H_2)^{-1} = I + H_1 - H_2 + {H_2}^2(I{+}H_2)^{-1}\\\\ P_2 &= H_2 + (I{+}H_1)^{-1} = I + H_2 - H_1 + {H_1}^2(I{+}H_1)^{-1}, \end{align} $$ so, since ${H_i}^2(I{+}H_i)^{-1}$ is positive semidefinite for $i=1,2$, we have $$ P_1{+}P_2 = 2I + {H_1}^2(I{+}H_1)^{-1} + {H_2}^2(I{+}H_2)^{-1} \ge 2I. $$ In particular, $P_1{+}P_2{-}2I\ge 0$, so this is a necessary inequality. Moreover, if $K\subset\mathbb{C}^N$ is the kernel of $P_1{+}P_2{-}2I$, then $K$ is also the set of vectors annihilated by both $H_1$ and $H_2$, so $P_1{-}I$ and $P_2{-}I$ must also vanish on $K$. In particular, if $W\subset\mathbb{C}^N$ is the orthogonal complement to $K$, then each $P_i$ preserves $W$, and we can restrict everything to $W\simeq\mathbb{C}^n$ for some $n\le N$, which reduces us to the case in which $P_1{+}P_2{-}2I > 0$, which, in turn, implies that $H_1{+}H_2>0$, so I will assume this from now on.

In fact, one can almost solve for $H_1$ and $H_2$. Set $M=(I{+}H_1)(I{+}H_2)$ and note that the equations imply $$ (I{+}P_1)(I{+}P_2) = 2I + M + M^{-1}. $$ Setting $Q = \tfrac12(P_1P_2{+}P_1{+}P_2-I)$, one gets $M{+}M^{-1} = 2Q$, so that $Q$ must commute with $M$, which allows the above equation to be written in the form $$ (M{-}Q)^2 = Q^2 - I. $$ In particular, $Q^2-I$ must be a square, which is a nontrivial condition on $Q$. (By the way, neither $M$ nor $Q$ is Hermitian symmetric, a priori.) If $Q^2-I$ has $n$ distinct nonzero eigenvalues (which is 'generic'), then it has exactly $2^n$ 'square roots'. If $Q^2-I$ has multiple eigenvalues, then it may have positive dimensional families of square roots. In any event, let $(Q^2{-}I)^{1/2}$ denote one of the square roots of $Q^2{-}I$. Then $M$ must satisfy $M = Q + (Q^2{-}I)^{1/2}$ for some one of these square roots. One then has $$ (I{+}P_1)(I{+}H_2) = M + I = I + Q + (Q^2{-}I)^{1/2}, $$ so $$ H_2 = (I{+}P_1)^{-1}\bigl(I + Q + (Q^2{-}I)^{1/2}\bigr) - I, $$ Now we see a necessary condition on the pair $(P_1,P_2)$, which is that the expression on the right hand side of the above equation must be a positive semidefinite matrix for some choice of the square root. One gets a similar condition when one solves for $H_1$. These are the necessary and sufficient conditions that characterize the image $F\bigl(\mathcal{K}_N\times\mathcal{K}_N\bigr)$.

I assume that the problem is to try to determine which pairs $(P_1,P_2)$ of positive definite Hermitian symmetric $N$-by-$N$ matrices can be written in the above form for some pair $(H_1,H_2)$ of positive semi-definite Hermitian symmetric $N$-by-$N$ matrices.

Let $\mathcal{H}_N$ denote the set of Hermitian symmetric $N$-by-$N$ matrices, and let $\mathcal{P}_N\subset \mathcal{H}_N$ denote the open convex cone consisting of the positive definite ones, with $\mathcal{K}_N = \overline{\mathcal{P}_N}$ denoting its closure, i.e., the set of positive semidefinite ones. Finally, let $\mathcal{J}_N\subset\mathcal{H}_N$ denote the convex open set consisting of the elements $H\in \mathcal{H}_N$ such that $I{+}H$ lies in $\mathcal{P}_N$. Note that $\mathcal{P}_N\subset \mathcal{K}_N\subset \mathcal{J}_N$. Consider the smooth map $F:\mathcal{J}_N\times \mathcal{J}_N\to \mathcal{P}_N\times \mathcal{P}_N$ defined by $$ F(H_1,H_2) = \bigl(H_1 + (I{+}H_2)^{-1}, H_2 + (I{+}H_1)^{-1}\bigr). $$ A straightforward computation shows that if $(H_1,H_2)\in\mathcal{K}_N\times\mathcal{K}_N$ satisfies $H_1{+}H_2\in \mathcal{P}_N$, then $F'(H_1,H_2):\mathcal{H}_N{\oplus}\mathcal{H}_N\to \mathcal{H}_N{\oplus}\mathcal{H}_N$ is an isomorphism, so, in particular, $F$ is a local diffeomorphism on a neighborhood of $(H_1,H_2)$. In particular, the mapping $$ F:\mathcal{P}_N\times \mathcal{P}_N\to \mathcal{P}_N\times \mathcal{P}_N $$ is a local diffeomorphism, so $F\bigl(\mathcal{P}_N\times\mathcal{P}_N\bigr)$ is an open subset of $\mathcal{P}_N\times \mathcal{P}_N$. Thus, there cannot be any algebraic relations implied between $P_1$ and $P_2$ in order that there exist an $(H_1,H_2)\in\mathcal{K}_N\times\mathcal{K}_N$ so that $F(H_1,H_2) = (P_1,P_2)$. The most one can hope for is inequalities relating $P_1$ and $P_2$ that might characterize the image $F\bigl(\mathcal{K}_N\times\mathcal{K}_N\bigr)$ in $\mathcal{P}_N\times\mathcal{P}_N$.

One inequality is very easy, and turns out to be quite useful: Note that $$ \begin{align} P_1 &= H_1 + (I{+}H_2)^{-1} = I + H_1 - H_2 + {H_2}^2(I{+}H_2)^{-1}\\\\ P_2 &= H_2 + (I{+}H_1)^{-1} = I + H_2 - H_1 + {H_1}^2(I{+}H_1)^{-1}, \end{align} $$ so, since ${H_i}^2(I{+}H_i)^{-1}$ is positive semidefinite for $i=1,2$, we have $$ P_1{+}P_2 = 2I + {H_1}^2(I{+}H_1)^{-1} + {H_2}^2(I{+}H_2)^{-1} \ge 2I. $$ In particular, $P_1{+}P_2{-}2I\ge 0$, so this is a necessary inequality. Moreover, if $K\subset\mathbb{C}^N$ is the kernel of $P_1{+}P_2{-}2I$, then $K$ is also the set of vectors annihilated by both $H_1$ and $H_2$, so $P_1{-}I$ and $P_2{-}I$ must also vanish on $K$. In particular, if $W\subset\mathbb{C}^N$ is the orthogonal complement to $K$, then each $P_i$ preserves $W$, and we can restrict everything to $W\simeq\mathbb{C}^n$ for some $n\le N$, which reduces us to the case in which $P_1{+}P_2{-}2I > 0$, which, in turn, implies that $H_1{+}H_2>0$, so I will assume this from now on.

In fact, one can almost solve for $H_1$ and $H_2$. Set $M=(I{+}H_1)(I{+}H_2)$ and note that the equations imply $$ (I{+}P_1)(I{+}P_2) = 2I + M + M^{-1}. $$ Setting $Q = \tfrac12(P_1P_2{+}P_1{+}P_2-I)$, one gets $M{+}M^{-1} = 2Q$, so that $Q$ must commute with $M$, which allows the above equation to be written in the form $$ (M{-}Q)^2 = Q^2 - I. $$ In particular, $Q^2-I$ must be a square, which is a nontrivial condition on $Q$. (By the way, neither $M$ nor $Q$ is Hermitian symmetric, a priori.) If $Q^2-I$ has $n$ distinct nonzero eigenvalues (which is 'generic'), then it has exactly $2^n$ 'square roots'. If $Q^2-I$ has multiple eigenvalues, then it may have positive dimensional families of square roots. In any event, let $(Q^2{-}I)^{1/2}$ denote one of the square roots of $Q^2{-}I$. Then $M$ must satisfy $M = Q + (Q^2{-}I)^{1/2}$ for some one of these square roots. One then has $$ (I{+}P_1)(I{+}H_2) = M + I = I + Q + (Q^2{-}I)^{1/2}, $$ so $$ H_2 = (I{+}P_1)^{-1}\bigl(I + Q + (Q^2{-}I)^{1/2}\bigr) - I = \tfrac12(P_2{-}I) + (I{+}P_1)^{-1}(Q^2{-}I)^{1/2}\ , $$ Now we see a necessary condition on the pair $(P_1,P_2)$, which is that the expression on the right hand side of the above equation must be a positive semidefinite matrix for some choice of the square root. One gets a similar condition when one solves for $H_1$. These are the necessary and sufficient conditions that characterize the image $F\bigl(\mathcal{K}_N\times\mathcal{K}_N\bigr)$.