Induced isomorphisms on cohomology of double differential complexes Let $K^{\bullet,\bullet}=\bigoplus_{p,q\ge0}K^{p,q}$ be a double differential complex, i.e. we have differential operators $$\cdots\stackrel{d}{\to} K^{p,q-1}\stackrel{d}{\to} K^{p,q}\stackrel{d}{\to} K^{p,q+1}\stackrel{d}{\to}\cdots $$  and $$\cdots\stackrel{\delta}{\to} K^{p-1,q}\stackrel{\delta}{\to} K^{p,q}\stackrel{\delta}{\to} K^{p+1,q}\stackrel{\delta}{\to}\cdots.$$
We may define a single differential complex 
$$K^\bullet=\bigoplus_n\left(\bigoplus_{p+q=n}K^{p,q}\right)$$
with differential operator $D=\delta+(-1)^pd$. Let $L^{\bullet,\bullet}$ be a second double complex, also with differential operators $d$ and $\delta$. Let $f:K^{\bullet,\bullet}\to L^{\bullet,\bullet}$ be a complex homomorphism in the following sense: $f$ is  a vector space homomorphism, $f(K^{p,q})\subset L^{p,q}$, and $f$ commutes with $d$ and $\delta$. 
Since $f$ commutes with $d$, it descends to a map $f^*:H_d(K^{p,\bullet})\to H_d(L^{p,\bullet})$ of cohomologies, where
$$K^{p,\bullet}=\bigoplus_q K^{p,q},$$
with differential $d$ and similarly for $L^{p,\bullet}$. Suppose that $f$ induces an isomorphism $H_d(K^{p,\bullet})\cong H_d(L^{p,\bullet})$ for all $p$ in this manner. Does it follow that this $f^*$ also gives an isomorphism between $H_D(K^\bullet)$ and $H_D(L^\bullet)$? 
I know this to be true when the $\delta$-cohomologies are zero, i.e. the $q$ rows are all exact. In Bott & Tu, this general case is claimed, but not proven. 

My earlier answer gives a counterexample that ignores your condition that $K$ and $L$ live in the first quadrant.  Given that condition, what you want is true:
Let $C^{\bullet,\bullet}$ be the mapping cone of $f$.  Then the long exact cohomology sequence for $f^{p,\bullet}$ shows that the horizontal cohomology of
$C^{\bullet,\bullet}$ is zero.  Therefore the first spectral sequence for the cohomology of $C^{\bullet,\bullet}$ collapses at $E^2$. But because $C^{\bullet,\bullet}$ lives in the first quadrant, the spectral sequence must converge, necessarily to zero, which gives your desired result.
