It is easy to see that within the disk algebra $A(D)$ $$\Delta(z):= \begin{pmatrix} 1&0\\z&1 \end{pmatrix}\; \begin{pmatrix} 1&1\\0&1 \end{pmatrix}= \begin{pmatrix} 1&1\\z&1+z \end{pmatrix} $$ is a product of two exponential matrices. Is $\Delta(z)$ itself an exponential matrix of a holomorphic matrix $M(z)$ ? I don't think so, but could not come up with a proof. Is there a general method to deal with these questions?

$\begingroup$ See again my answer, the upperleft entries lacked the correct sign. $\endgroup$ – Denis Serre May 2 '16 at 12:38
$\Delta(z)$ does not have a nonpositive real eigenvalue for $z < 4$, so the principal branch of the logarithm is defined and analytic on a neighbourhood of its spectrum, and thus the holomorphic functional calculus produces the desired analytic logarithm $M(z)$ for $z < 4$.
It seems that $\Delta(z)$ is the exponential of an holomorphic $M(z)$. Using the eigenvalues and eigenvectors of $\Delta$, I find $$M(z)=\frac\mu{\sqrt{z(z+4)}}\begin{pmatrix} z & 2 \\ 2z & z \end{pmatrix},$$ where $\mu$ is the solution of $$\sinh\mu=\frac12\sqrt{z(z+4)}.$$ Because $\sinh^{1}$ is an odd holomorphic function, the quotient $\frac\mu{\sqrt{z(z+4)}}$ can be chosen holomorphic, at least in a domain where $$\left\frac14z(z+4)\right<1,$$ because the convergence radius of the power series $$\sinh^{1} x=x\frac1{2\cdot3}x^3+\frac{1\cdot3}{2\cdot4\cdot5}x^5+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot7}x^7+\cdots$$ equals $1$. To have a positive answer to the question, we need to prove that $\sinh^{1}$ extends holomorphically to a slightly larger domain, containing the image of $D$ under the multivalued map $$z\mapsto\frac12\sqrt{z(z+4)}\,.$$ A good news is that this image does not reach the singularities $\pm i$ of the series above.
To address the part of the question about a general method  seems like the BakerCampbellHausdorff formula can give something tractable in this case. Denoting $e=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $zf=\left(\begin{smallmatrix}0&0\\z&0\end{smallmatrix}\right)$ (so that $\Delta=\operatorname{Exp}(e).\operatorname{Exp}(zf)$), one has $$ zh:=[e,zf]=z(e.ff.e)=\left(\begin{smallmatrix}z&0\\0&z\end{smallmatrix}\right) $$ so that the whole Lie algebra generated by $e$ and $zf$ comes out the vector space spanned by $e_n=(2z)^{n1}e$, $f_n=2^{n1}z^nf$, $h_n=2^{n1}z^nh$, $n=1,2,...$, with brackets $$ [e_m,e_n]=[f_m,f_n]=[h_m,h_n]=0, $$ $$ [h_m,e_n]=2e_{m+n}, $$ $$ [h_m,f_n]=2f_{m+n} $$ and $$ [e_m,f_n]=h_{m+n1}. $$ In other words the Lie algebra is (almost all of the) $\mathfrak{sl}_2[z]$, its exponential map must be well studied by physicists.
In fact if I am not mistaken the BCH formula gives in this case \begin{multline*} \Delta=\operatorname{Exp}\left(e_1+f_1\frac12h_1\frac1{12}(e_2+f_2\frac12h_2)+\frac1{120}(e_3+f_3\frac12h_3)+...\right.\\\left....+(1)^{n1}\frac{(n1)!}{4^{n1}(2n1)!!}(e_n+f_n\frac12h_n)+...\right), \end{multline*} and since $$ e_n+f_n\frac12h_n=\frac{(2z)^{n1}}2\begin{pmatrix}z&2\\2z&z\end{pmatrix}, $$ we get $$ M(z)=\frac12\sum_{n=1}^\infty(1)^{n1}\frac{(n1)!}{(2 n1)\text{!!}}\left(\frac z2\right)^{n1}\begin{pmatrix}z&2\\2z&z\end{pmatrix} $$ which (I think) boils down to the same $M(z)$ as in another answer.