The question is regarding positivity of bounded operators on Hilbert spaces. I try to briefly clarify a part of my motivation below.
It is direct to see that if $P,Q\in\mathcal{B}(\mathcal{H})$ are positive, then $PQ$ is positive iff $P$ and $Q$ commute (here is a link to another answer that deals with this). The following question is perhaps interesting: if $p,q$ are two non-normal commuting elements of $\mathcal{B}(\mathcal{H})$, then for positive operators $P=p^\star p$ and $Q=q^\star q$, is $PQ$ positive (equivalent to asking if they commute)? It turns out that the answer is NO! It is easy to obtain an example even in $2 \times 2$ matrices: $p = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $q = \begin{bmatrix} 1 & i \\ 0 & 1 \end{bmatrix}$ yield a counterexample, where $pq=qp$ but $PQ=p^\star p q^\star q \neq QP$, and isn't positive. Of course, it isn't even hermitian. However, somewhat naively, we observe that the operator obtained by moving all the adjoints to the left (in any order, as $p$ and $q$ commute), yields $q^\star p^\star p q$, which is positive.
Now, suppose we consider the following problem: $T_1 , T_2$ be a pair of non-normal commuting contractions on a Hilbert space (non-normal, so spectral theorem based techniques probably wouldn't come in handy). Then, the corresponding defect operators, $D_i = \sqrt{\mathbb{1}-T_i ^\star T_i}$, $i=1,2$, are positive contractions, but do not commute in general. Hence, $D_1 D_2 = (\mathbb{1}-T_1 ^\star T_1)(\mathbb{1}-T_2 ^\star T_2)=\mathbb{1}-T_1 ^\star T_1 -T_2 ^\star T_2 + T_1 ^\star T_1 T_2 ^\star T_2$ cannot be expected to be a positive operator in general (not even hermitian). However, if we consider the corresponding operator obtained by the naive recipe described above (moving all adjoints to the left), $\mathbb{1}-T_1 ^\star T_1-T_2^ \star T_2+T_1 ^\star T_2^\star T_1 T_2$, it is obviously hermitian, but is it positive?
One of my failed attempts I should probably mention: observe that $\mathbb{1}-T_1 ^\star T_1-T_2^ \star T_2+T_1 ^\star T_2^\star T_1 T_2 = (\mathbb{1}-T_1 ^\star T_1)-T_2^ \star(\mathbb{1}-T_1 ^\star T_1) T_2 = (\mathbb{1}-T_2 ^\star T_2)-T_1^ \star(\mathbb{1}-T_2 ^\star T_2) T_1$. I try to solve a related problem: for positive contractions $a$ and $b$, would $a-b^\star a b$ be positive? I found a (counter)example (again, for $2\times 2$ matrices) for which the answer to this question in in the negative, but failed to find a contraction $T$ such that $T^\star T = \mathbb{1} – a$ and and $bT=Tb$, so probably this is inconclusive regarding the original question.
I was also wondering if Ando's Theorem would help, but failed to see how.
A little bit on my background, in case it is relevant to the question: I studied, among other things, the basics of C* Algebras, positive and CP maps, and dilations, for my Masters' thesis.
Best Answer
The answer is negative. Here is a counter example: let $ T_1={1\over \sqrt 2}\pmatrix{1 & 0 \cr 1 & 0}, $ so that $$ 1-T_1 ^* T_1 = \pmatrix{0 & 0 \cr 0 & 1}. $$ The matrices commuting with $T_1$ are of the form $T_2=\pmatrix{a & 0 \cr b & a-b}$ so let us take any such matrix, say with real entries. Using your identity we obtain $$ {1}-T_1 ^* T_1-T_2^* T_2+T_1 ^* T_2^* T_1 T_2 = ({1}-T_1 ^* T_1)-T_2^*({1}-T_1 ^* T_1) T_2 = $$$$ \pmatrix{0 & 0 \cr 0 & 1} -T_2^* \pmatrix{0 & 0 \cr 0 & 1} T_2 = \pmatrix{0 & 0 \cr 0 & 1} - \pmatrix{b^2 & b(a-b)\cr b(a-b) & (a-b)^2} = $$$$ \pmatrix{-b^2 & -b(a-b)\cr -b(a-b) & 1-(a-b)^2} $$ which is not positive in case $b≠0$. It remains to make $T_2$ a contraction but this is easy to do by choosing small values of $a$ and $b$.