Calculating the exponential of a $4 \times 4$ matrix

Consider $M(t) = \exp(t A)$, and as you noticed, it has block-diagonal form $$ M(t) = \left(\begin{array}{cc} \exp(t B) & n(t) \\ 0_{2 \times 2} & \exp(t B) \end{array} \right). $$ Notice that $M^\prime(t) = A \cdot M(t)$, and this results in a the following differential equation for $n(t)$ matrix: $$ n^\prime(t) = \mathbb{I}_{2 \times 2} \cdot \exp(t B) + B \cdot n(t) $$ which translates into $$ \frac{\mathrm{d}}{\mathrm{d} t} \left( \exp(-t B) n(t) \right) = \mathbb{I}_{2 \times 2} $$ which is to say that $n(t) = t \exp(t B)$.

A different, but rather specific, strategy would be to use the ring homomorphism $${a+bi\in\mathbb C \mapsto \pmatrix{a&-b \\ b&a}\in\mathbb R^{2\times 2}}$$in the block decomposition. Then your problem is equivalent to finding $$e^{t\pmatrix{1+i & 1\\ 0 & 1+i}}=e^{\pmatrix{t+ti & t\\ 0 & t+ti}}=e^{t+ti}e^{\pmatrix{0&t\\0&0}}=(e^{t+ti})\pmatrix{1&t\\0&1}$$ which unfolds to $$\pmatrix{e^t\cos t & -e^t\sin t & t e^t \cos t & -t e^t \sin t \\ e^t \sin t & e^t \cos t & t e^t \sin t & t e^t \cos t \\ 0 & 0 & e^t\cos t & -e^t\sin t \\ 0&0& e^t\sin t & e^t\cos t }$$